Write $f(z) = u(r, \theta) + iv(r, \theta)$; suppose that the first-order partials of $u, v$ with respect to $r, \theta$ are continuously differentiable in some neighborhood of $z$ and satisfy $ru_r = v_\theta$ and $rv_r = -u_\theta$. Then, $f$ is differentiable at $z$.
One approach is to show that $u_x, u_y, v_x, v_y$ are continuously differentiable and satisfy $u_x = v_y$ and $-v_x = u_y$. To that end, it is necessary to show that $\frac{\partial r}{\partial x}, \frac{\partial \theta}{\partial x} \ldots$ are continuously differentiable. Then, via the chain rule, $u_x = u_r \frac{\partial r}{\partial x} + u_\theta \frac{\partial \theta}{\partial x} \ldots$ so on and so forth.
But if you are defining $\theta$ with $\tan^{-1}\frac{y}{x}$, then you will always have a branch cut somewhere. So, this approach will not work when we're at the branch cut. How do you deal with this?
Otherwise, to do it via definition of differentiability, then you need to untangle the nightmare in $\triangle z = (r + \triangle r)e^{i (\theta + \triangle \theta)} - re^{i \theta}$.
