Reading the question and the comments, I believe first of all some clarification is needed.
First of all you need to tell us where you intend to calculate the Thévenin equivalent: you can do that across any two terminals of that particular circuit since it is linear and the controlled voltage source control quantity is inside the circuit you want to reduce.
For example, if you want to calculate the Thévenin equivalent impedance of the net across the \$ V_{rms} \$ voltage source it would be zero since it is ideal.
To calculate the Thévenin equivalent impedance across any two other terminals you first need to turn off any independent voltage/current source, then connect a test source \$V_t\$ across the terminal you have chosen, solve the circuit, obtain the test current \$I_t\$ and finally compute \$Z_{th}=\frac{V_t}{I_p}\$.
I believe that what you want to do is calculate the Thévenin impedance seen from \$V_{rms}\$. To do that \$V_{rms}\$ must be removed, then usual circuit approach can be applied.
Let's call N the node where the resistor, the inductor and the capacitor are connected. From \$R_1\$ arrives a current \$I_t\$ that then splits in the two branches. Let's call \$I^*\$ the current that flows in \$L_1\$, and \$V_n\$ the voltage at node n. Let's finally tie the node \$V_{rms}\$, \$C_1\$, \$Vccs_1\$ to the ground.
If \$Z_n\$ is the impedance seen across node n and ground removing \$R_1\$, you can write \$Z_{th} = R_1 + Z_n\$. Can we calculate \$Z_n\$ easily? Yes.
You can write:
$$
I^* = I_t - \frac{V^*}{Z_{C_1}}
$$
$$
I^* = \frac{V^*-\frac{I_t}{2}}{Z_{L_1}}
$$
From these you can derive:
$$
V^* = I_t\frac{1+\frac{1}{2Z_{L_1}}}{Z_{L_1}//Z_{C_1}}
$$
where \$Z_{L_1}//Z_{C_1} = \frac{Z_{L_1}Z_{C_1}}{Z_{L_1}+Z_{C_1}}\$
Now, \$Z_n = \frac{V_n}{I_t}=\frac{1+\frac{1}{2Z_{L_1}}}{Z_{L_1}//Z_{C_1}}\$, and finally:
$$
Z_{th} = R_1 + \frac{V_n}{I_t}=\frac{1+\frac{1}{2Z_{L_1}}}{Z_{L_1}//Z_{C_1}}=(1.9375-0.24j)\Omega
$$
Say you connect a voltage source Va at the external terminals across which you want to find the the equivalent resistance. Now if you change the voltage Va, voltage across the elements in the network changes, so does the current. The ratio of this change in voltage to that of current, \$\Delta V/\Delta I\$ is the equivalent resistance offered by that element.
The total resistance of a network can be calculated by replacing each element by its equivalent resistance.
For a reistor, \$\Delta V/\Delta I\$ will be its resistance value only.
Since voltage across an independent voltage won't change (\$\Delta V=0\$), its equivalent resistance will be zero. Hence it can be replaced by a short. Similarly, independent current source (\$\Delta I=0\$) can be replaced by an open circuit. Which is basically 'switching off' these sources.
But for dependent sources, this \$\Delta V/\Delta I\$ ratio need not be zero or infinity as the value of source depends on voltage/current at some other node/branch in the network. Hence effectively it can have a non zero finite resistance to offer and hence can not be 'switched off'.
Best Answer
It's entirely possible for both the short circuit current and open-circuit voltage to be zero. For example, if the network consists of a single resistor (or probably any linear network with no independent sources).
In this case you just need to use a different pair of "test points" to find the equivalent resistance. For example, you can use V=0 (the short-circuit case) and V = 1 V. Then \$R_{th}=\frac{1}{I(V=1) - I(V=0)}\$, or more generally \$R_{th}=\frac{V_1-V_2}{I(V_1) - I(V_2)}\$.
Basically one way to state Thevenin's theorem is that the I-V characteristic of any one-port network of independent sources and linear elements will be a straight line. And \$R_{th}\$ is the slope of that line, \$\dfrac{\mathrm{d}v}{\mathrm{d}i}\$. Just as with any other linear function, you can can use any two points to find the slope of the line as rise/run.
Edit
You've mis-stated this a bit (where I added emphasis).
The usual way to find the Thevenin or Norton equivalent circuit is to find \$V_{OC}\$, the voltage across an open-circuit on the output, and \$I_{SC}\$, the current through a short-circuit on the output.
It happens that \$V_{OC}\$ is also the Thevenin-equivalent voltage source value and \$I_{SC}\$ is the Norton-equivalent current source value. But when you're measuring an arbitrary network to find the equivalents, you should think of these as the open-circuit output voltage and short-circuit output current.
I believe \$V_{OC}\$ and \$I_{SC}\$ will both be zero whenever there are no non-zero independent sources in the network being modeled. However this does not mean that the Thevenin resistance is indeterminate, as I explained above.
You could also contrive a case where there's a non-zero independent source and a dependent source that directly opposes it, like for example,
simulate this circuit – Schematic created using CircuitLab
I think this is a different question. The Thevenin equivalent resistance will only be indeterminate if it goes to infinity, for example when the network being modeled is an ideal current source.