Solution

For the first circuit we have:

\begin{equation} \mathcal{E}+\mathcal{E}_{i1}=IR \end{equation}

where $\mathcal{E}_{i1}$ is an induced emf of a motor in the first case.

Similarly,for the second one we could write:

\begin{equation} \mathcal{E}_0+\mathcal{E}_{i2}=0 \end{equation}

where $\mathcal{E}_{i2}$ is an induced emf of a motor in the second circuit.

Using the fact the induced emf is proportional to its frequency we get:

\begin{equation} \frac{\mathcal{E}_{i1}}{f}=\frac{\mathcal{E}_{i2}}{f_0} \end{equation}

Or it is the same as:
\begin{equation} \mathcal{E}_{i1}=-\mathcal{E}_0\frac{f}{f_0} \end{equation}

In the steady state, we know that the net torque is zero,so:

\begin{equation} M+M_m=0 \end{equation}

where $M_m$ is an electromagnetic torque of the motor which we could through it's angular frequency and power $P_m$:

\begin{equation} \omega M_m=P_m \end{equation}

But we could write $P_m$ as $\mathcal{E}_{i1}I$ and $\omega$ as $2\pi f$ for the first case.By plugging them into eq(6) and using eq(5) we get $I$:

\begin{equation} I=\frac{2\pi M f_0}{\mathcal{E}_0} \end{equation}

Finally,plugging last equation into the first one and using eq(4) we get our answer for frequency:

\begin{equation} f=f_0(\frac{\mathcal{E}}{\mathcal{E}_0}-\frac{2\pi MRf_0}{\mathcal{E}_0^2}) \end{equation}