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.
No, this does not work this way as pointed out by another fellow in a different post of yours: the first \$RC\$ section is loaded by the Sallen-Key filter input impedance so you can't neglect it. You could if you would buffer the first network before driving the second filter. Another option would be to start on the left by the Sallen-Key structure then feeding the \$RC\$ filter. Considering a 0-\$\Omega\$ output impedance for the op amp then you could cascade (multiply) the transfer functions.
I've shown here without writing a single line of algebra how you can determine this transfer function using fast analytical circuits technique described here. Considering well-separated real poles, the transfer function can be approximated as
and if you plot this expression, you have
Edit: as correctly underlined by LvW, the above arrangement does not reflect a Bessel transfer function but rather a series of 3 cascaded poles. The transfer function describing this network actually obeys the following expression: \$H(s)=\frac{1}{1+b_1s+b_2s^2+b_3s^3}\$ in which
\$b_1=R_1C_1+(R_1+R_2+R_3)C_2\$
\$b_2=C_2(C_1R_1(R_2+R_3)+C_3R_3(R_1+R_2))\$
\$b_3=C_1C_2C_3R_1R_2R_3\$
The transfer function of a 3rd-order Bessel low-pass filter normalized at a 1-rad/s characteristic angular frequency is given by \$H(s)=\frac{1}{1+s+s^2\frac{6}{15}+\frac{s^3}{15}}\$. This expression can be refactored in a more friendly format, as its roots comprise 1 real pole and 2 complex poles: \$H(s)\approx\frac{1}{(1+0.43s)(1+0.57s+0.155s^2)}\$
Assume we would like to tune this filter to 1 kHz, then you have to scale the formula according to:
\$H(s)=\frac{1}{1+s\frac{1}{k}+s^2\frac{0.4}{k^2}+s^3\frac{0.067}{k^3}}\$
which is approximately equal to:
\$H(s)\approx\frac{1}{(1+\frac{0.43}{k}s)(1+\frac{0.57}{k}s+\frac{0.155}{k^2}s^2)}\$
with \$k=2\pi 1000=6.283\times 10^3\$
I can plot these two expressions and they give identical dynamic responses in magnitude and phase:
Now, you have to determine the component values so that the coefficients determined by the component combinations leads to:
\$b_1=159\;ms\$, \$b_2=1.013\times 10^4\;µs^2\$ and \$b_3=2.688\times 10^{-13}s^3\$
This is a system of 3 equations/3 unknown if you fix \$C_1\$, \$C_2\$ and \$C_3\$ for instance. If you use the tool here, then you have the following component values for a 1-kHz characteristic frequency:
\$R_1=9.1\;k\Omega\$ \$R_2=91\;k\Omega\$ \$R_3=36\;k\Omega\$ \$C_1=6.8\;nF\$ \$C_2=680\;pF\$ \$C_3=1.8\;nF\$
If now plot the expression that I determined here and featuring the above component values versus the response of the 3rd-order Bessel transfer function we have derived, the plots are identical:
Best Answer
The study of this simple filter can be undertaken applying the current division law and then you develop and rearrange the expression:
Or you apply the fast analytical circuits techniques or FACTs and obtain the result in a few minutes, the time to write this answer: reduce the excitation to 0 A (open-circuit the left-side source) and "look" at the resistance offered by the connecting terminals of the capacitor and the inductor. Collect the time constants in this mode, assemble them to form the denominator and there you go for this simple circuit:
Considering a low quality factor, you can even apply the low-\$Q\$ approximation and obtain a new expression with two cascaded poles. The resulting transfer function plot is there:
No need for KVL or KCL, the FACTs get you to the well-ordered transfer function by inspecting the circuit without writing a line of algebra.