= Solution
Apply the <multivariable chain rule>, keeping the other new coordinate fixed in each differentiation. At fixed $\theta$, the <derivatives> of $x,y$ with respect to $r$ are $\cos2\theta,\sin2\theta$. At fixed $r$, their <derivatives> with respect to $\theta$ are $-2r\sin2\theta,2r\cos2\theta$. Thus
$$
\boxed{\left(\frac{\partial f}{\partial r}\right)_\theta
=\cos2\theta\left(\frac{\partial f}{\partial x}\right)_y
+\sin2\theta\left(\frac{\partial f}{\partial y}\right)_x},
$$
$$
\boxed{\left(\frac{\partial f}{\partial\theta}\right)_r
=-2r\sin2\theta\left(\frac{\partial f}{\partial x}\right)_y
+2r\cos2\theta\left(\frac{\partial f}{\partial y}\right)_x}.
$$
The factor two arises because the ordinary polar angle is $2\theta$.
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