Find the frequency at which Vin gets to Vo without the phase shift
As correctly suggested by sstobbe, the way to go is to determine the impedance offered by node X in your schematic. The easiest path is using the FACTs and isolate the portion at node X first. This is what is done below:



The input resistance \$R_0\$ when \$C_1\$ is open circuited is equal to \$R_1\$. This is because the open cap. blocks the dc bias and you have 0 V across \$R_1\$ and it propagates to the right side of \$R_2\$ which has the same value as \$R_1\$. To obtain the pole, simply reduce the excitation to 0 A (open circuit the current test generator) and "look" at the resistance offered by the capacitor connections in this mode (see below):



If you do the maths ok, you should find a resistance equal to \$R_1(1-(\frac{A_{OL}}{1+A_{OL}})^2)\$ leading to a pole placed at \$\omega_p=\frac{1}{\tau_1}\$ with \$\tau_1=C_1R_1(1-(\frac{A_{OL}}{1+A_{OL}})^2)\$. With perfect op amps, this pole is extremely high and can be neglected.

To obtain the zero, null the response across the first test generator. This is a degenerate case in which you now "look" at the resistance offered by \$C_1\$'s terminals while the current source is shorted. See below:



If you observe this circuit correctly, you realize that the resistance in this case is simply \$R_1\$ or \$R\$ in your case. The zero is then located at \$\omega_z=\frac{1}{\tau_2}\$ with \$\tau_2=R_1C_1\$.

The complete transfer function of the impedance offered by node X is then \$Z_X(s)=R_0\frac{1+\frac{s}{\omega_z}}{1+\frac{s}{\omega_p}}\approx R_0(1+\frac{s}{\omega_z})\$. Now neglecting the 1 contribution for high \$s\$ values, we have \$Z_X(s)\approx s\frac{R}{\omega_z}\$. This leads to an inductance value \$L=R^2C\$. With the values given in the circuit I built, the inductor is 100 H. See the below Mathcad for the details:



You can now rework you original circuit by inserting the inductor in the circuit:



However, I will leave to you the determination of the resonant frequency of this 2nd-order circuit. You can use the FACTs if you want : )