Your calculation of the impedance seen by the source is correct.
Clearly, there is a 'special' (angular) frequency
$$\omega_0 = \frac{1}{\sqrt{LC}}$$
where there is a pole in the impedance - the impedance goes to infinity.
Now, let's look at the dual of the circuit given:
simulate this circuit – Schematic created using CircuitLab
For the dual circuit, the impedance seen by the source is
$$Z = R||(j\omega L + \frac{1}{j \omega C}) = R \frac{1 - \omega^2LC}{1 - \omega^2LC + j\omega RC} $$
and now we have a zero at \$\omega_0\$ - the impedance goes to zero.
In both of these cases, the pole or zero is on the \$j \omega\$ axis. Generally, they are not.
so how do you find the resonance in general?
In this context (RLC), the resonance frequency is the frequency where the impedance of the inductor and capacitor are equal in magnitude and opposite in sign.
Update to address comment and question edit.
From the Wikipedia article "RLC circuit", "Natural frequency" section:
The resonance frequency is defined in terms of the impedance presented
to a driving source. It is still possible for the circuit to carry on
oscillating (for a time) after the driving source has been removed or
it is subjected to a step in voltage (including a step down to zero).
This is similar to the way that a tuning fork will carry on ringing
after it has been struck, and the effect is often called ringing. This
effect is the peak natural resonance frequency of the circuit and in
general is not exactly the same as the driven resonance frequency,
although the two will usually be quite close to each other. Various
terms are used by different authors to distinguish the two, but
resonance frequency unqualified usually means the driven resonance
frequency. The driven frequency may be called the undamped resonance
frequency or undamped natural frequency and the peak frequency may be
called the damped resonance frequency or the damped natural frequency.
The reason for this terminology is that the driven resonance frequency
in a series or parallel resonant circuit has the value1
$$\omega_0 = \frac {1}{\sqrt {LC}}$$
This is exactly the same as the resonance frequency of an LC circuit,
that is, one with no resistor present, that is, it is the same as a
circuit in which there is no damping, hence undamped resonance
frequency. The peak resonance frequency, on the other hand, depends on
the value of the resistor and is described as the damped resonance
frequency. A highly damped circuit will fail to resonate at all when
not driven. A circuit with a value of resistor that causes it to be
just on the edge of ringing is called critically damped. Either side
of critically damped are described as underdamped (ringing happens)
and overdamped (ringing is suppressed).
Circuits with topologies more complex than straightforward series or
parallel (some examples described later in the article) have a driven
resonance frequency that deviates from \$\omega_0 = \frac
{1}{\sqrt {LC}}\$ and for those the undamped resonance frequency, damped
resonance frequency and driven resonance frequency can all be
different.
See the "Other configurations" section for your 2nd circuit.
In summary, the frequencies at which the impedance is real, at which the impedance is stationary (max or min), and at which the reactances of the L & C are equal can be the same or different and each is some type of resonance frequency.
EDIT: Thanks to hryghr I see that the starting assumptions were incorrect. The transfer function magnitude can't be found that simply.
It is more than ten years since I considered my skills sharp on this
topic, and knives don't get sharper in the drawer! But I can't have
that I posted something formally incorrect, so here goes attempt #2:
I will derive the transfer function the dirty way .. using Kirchoff's
Current Law (KCL) (a very generic method). I call the output node \$V_{o}\$, and the middle node \$V_{x}\$. For the following equations i cut down on writing by
writing \$V_{o}\$ instead of the more accurate \$V_{o}(s)\$ :
I: KCL in \$V_{o}\$:
$$
\frac{V_{o}-V_{x}}{R_{2}}+sC_{2}V_{o}=0
$$
$$
V_{x}=V_{o}(1+sR_{2}C_{2})
$$
II: KCL in \$V_{x}\$:
$$
\frac{V_{x}-V_{i}}{R_{1}}+\frac{V_{x}-V_{o}}{R_{2}}+sC_{1}V_{x}=0
$$
Rearranging terms:
$$
R_{2}(V_{x}-V_{i})+R_{1}(V_{x}-V_{o})+sR_{1}R_{2}C_{1}V_{x}=0
$$
Rearranging terms:
$$
V_{x}(R_{1}+R_{2}+sR_{1}R_{2}C_{1})-R_{2}V_{i}-R_{1}V_{o}=0
$$
Substituting \$V_{x}\$ with result of I:
$$
V_{o}(1+sR_{2}C_{2})(R_{1}+R_{2}+sR_{1}R_{2}C_{1})-R_{2}V_{i}-R_{1}V_{o}+sR_{1}R_{2}C_{1}V_{o}=0
$$
Collecting terms for \$V_{o}\$
$$
V_{o}((1+sR_{2}C_{2})(R_{1}+R_{2}+sR_{1}R_{2}C_{1})-R_{1})=R_{2}V_{i}
$$
Rearranging:
$$
\frac{V_{o}}{V_{i}}=\frac{R_{2}}{(1+sR_{2}C_{2})(R_{1}+R_{2}+sR_{1}R_{2}C_{1})-R_{1}}
$$
Expanding terms:
$$
\frac{V_{o}}{V_{i}}=\frac{R_{2}}{R_{1}+R_{2}+sR_{1}R_{2}C_{1}+sR_{1}R_{2}C_{2}+sR_{2}^{2}C_{2}+s^{2}R_{1}R_{2}^{2}C_{1}C_{2}-R_{1}}
$$
\$R_{1}\$ cancels, then divide by \$R_{2}\$ top and bottom:
$$
\frac{V_{o}}{V_{i}}=\frac{1}{1+sR_{1}C_{1}+sR_{1}C_{2}+sR_{2}C_{2}+s^{2}R_{1}R_{2}C_{1}C_{2}}
$$
Prettified, the transfer function is:
$$
H(s)=\frac{V_{o}(s)}{V_{i}(s)}=\frac{1}{s^{2}R_{1}R_{2}C_{1}C_{2}+s(R_{1}C_{1}+R_{1}C_{2}+R_{2}C_{2})+1}
$$
This is probably a nice place to start converting to the standard form that
hryghr mentions. It may be that the corner frequency asked for relates to that form.
I won't bother to much with that, but move on to find the -3dB point.
The magnitude of the transfer function can for instance be found by
calculating:
$$
\left|H(\omega)\right|=\sqrt{H(s\rightarrow j\omega)H(s\rightarrow-j\omega)}
$$
Setting \$A=R_{1}R_{2}C_{1}C_{2}\$ and \$B=(R_{1}C_{1}+R_{1}C_{2}+R_{2}C_{2})\$
to simplify this calculation:
$$
\left|H(\omega)\right|=\frac{1}{\sqrt{((j\omega)^{2}A+(j\omega)B+1)((-j\omega)^{2}A+(-j\omega)B+1)}}
$$
$$
\left|H(\omega)\right|=\frac{1}{\sqrt{(-\omega{}^{2}A+j\omega B+1)(-\omega{}^{2}A-j\omega B+1)}}
$$
$$
\left|H(\omega)\right|=\frac{1}{\sqrt{\omega{}^{4}A^{2}-\omega{}^{2}A(j\omega B-j\omega B+1+1)+\omega^{2}B^{2}+(j\omega B-j\omega B)+1}}
$$
$$
\left|H(\omega)\right|=\frac{1}{\sqrt{\omega{}^{4}A^{2}+\omega{}^{2}(B^{2}-2A)+1}}
$$
Finding \$B^{2}-2A\$ gives you something like:
$$
R_{1}^{2}(C_{1}+C_{2})^{2}+C_{2}^{2}(2R_{1}R_{2}+R_{2}^{2})
$$
Then to find the -3dB point start at:
$$
\frac{1}{\sqrt{2}}=\frac{1}{\sqrt{\omega{}^{4}A^{2}+\omega{}^{2}(B^{2}-2A)+1}}
$$
$$
2=\omega{}^{4}A^{2}+\omega{}^{2}(B^{2}-2A)+1
$$
So far I have done it all by hand (hopefully no mistakes), but here
I call it a day, try mathematica, and get \$\omega\$ for the -3dB frequency as:
$$
w\to\sqrt{\frac{1}{A}-\frac{B^{2}}{2A^{2}}+\frac{\sqrt{8A^{2}-4AB^{2}+B^{4}}}{2A^{2}}}
$$
Best Answer
Your filter is an ideal resonant circuit - practically non-existent! Components, especially inductors are lossy and signal sources also have something, typically a series resistance. Finally there's a load in the output. It also takes something. The resulted circuit no more has a pole (=zero denominator) at the resonant frequency. If there's low enough resistance load resistor, this circuit can be a 2nd order low pass filter.
This filter pulls the signal source output on the knees at the resonant frequency if there's some resistance in the source and the output is unloaded. It's extremely important to understand it qualitatively - the LC series impedance is lowest at the resonant frequency and the resistance of the signal source makes with it a frequency dependent voltage divider. You can calculate what's left at the output, but you can see it also by simulating.
You should do some simulations to see the effect of the various resistances. Even this site has a good enough simulator.
Calculations with equations really are complex, but doable. Advanced filter designers know some shortcuts how to use already calculated filter tables, how to transform them to other frequencies and how to change component values for different signal source and load resistances. The theory has been there already about 80 years. It's packed also in filter design software and printed to handbooks.
For teaching you maybe should show results and the methods qualitatively. Don't alienate people with heavy equation manipulations. More likely let them find something wanted with simulations.
If transfer function manipulation with complex numbers is the main subject, then simulation can be useful to reveal errors.
You can calculate and plot transfer functions with Excel, but simulator can show the frequency response of a circuit without forcing one to write complex equations which are error prone.
Time domain analysis with simulators, Excel and formally with Laplace transforms can be interesting, but have you time for it? You know. My opinion is that you should show with simulations that too high resonant peak in the frequency response causes too much unwanted ringing when the signal has pulses.