I have looked at this filter and derived its transfer function using the fast analytical techniques or FACTs as described in the book I wrote on the subject. The technique consists of looking at the time constants involving each of the capacitors in two different conditions: a zeroed excitation (the input voltage source is replaced by a short circuit) and a nulled response. As you have two capacitors in the circuit with independent state variables, this is a second-order filter.
I used SPICE to verify all my calculations in Mathcad and a dc operating point is fast and efficient as the below figure shows. In this sheet, I have determined the time constants for the denominator (the input is zeroed for this purpose) and the numerator also:
I then assembled the time constants and formed second-order polynomials in the numerator and the denominator. If the capacitors have equal values, then you can define the resonant frequency \$\omega_0\$ as follows:
\$\omega_0=\frac{1}{C\sqrt{{R_2}{R_3}}}\$ with the labels shown in the sheet below:
A quick SPICE simulation lets me test the calculations and verify the notch attenuates by 32.25 dB and is tuned at 60 Hz:
Addition 1:
The difficulty here is to check this expression versus a brute-force formula and verify both curves perfectly superimpose. You have several options to determine the transfer function without resorting to the FACTs and I like superposition a lot because I can often inspect the circuit without writing equations. Here I will turn off the controlled source as recommended by the late Marshall Leach in his paper. The trick when using this approach is to first determine the control variable (voltage at node A here) then proceed with superposition. The below shot shows how to apply this to the circuit:
To verify this arm-long expression is exactly the same as the one derived with the FACTs, I subtracted magnitudes and phases, implying a result as low as the solver resolution which is a few picos. Any significant deviation indicates an error somewhere. As the below plots show, the agreement is perfect in magnitude and phase:
Addition 2:
I am a power supply designer more than a filter master :-) and I've read that the circuit connected at the low-side terminal of \$C_1\$ could be modeled as an emulated inductance. I derived the transfer function (impedance) and if you neglect the high-frequency pole of the impedance "seen" below \$C_1\$ low-side terminal, you indeed have an equivalent inductance defined as \$L_{eq}\approx R_3R_2C_2 \$ in series with resistance \$R_3\$:
The new transfer function is thus that of an impedance divider involving \$R_1\$, \$C_1\$ and the equivalent lossy inductance. The transfer function is not that far from the original full-blown expression:
This time, this is it for the reply!
