This is the work that I've done.
The transformation between polar and rectangular coordinates can be expressed as: $x = r \cos \theta$, $y = r \sin \theta$.
Hence, $\frac{\partial x}{\partial r} = \cos \theta$.
Furthermore, we can write $r = \sqrt{x^2 + y^2}$ as part of the reverse transformation. Taking the partial derivative of this with respect to $x$, we have $\frac{\partial r}{\partial x} = \frac{x}{\sqrt{x^2 + y^2}} = \frac{x}{r} = \frac{r \cos \theta}{r} = \cos \theta$.
Have I made some mistake or is it correct that both of these partial derivatives are equal to $\cos \theta$? How can one interpret this fact?
Thank you for any insight!
Unfortunately, in the derivation of $\frac{\partial r}{\partial x},$ it does not make sense to substitute $x = r \cos \theta$ because you are viewing $r$ and $x$ as functions of $x$ and $y.$ In fact, it is actually true that for fixed $\theta,$ we have that$$\frac{\partial x}{\partial r} = \frac{1}{\frac{\partial r}{\partial x}}.$$
Proof. Using implicit differentiation on $x = r \cos \theta,$ we have that $$1 = \frac{\partial}{\partial x} x = \frac{\partial}{\partial x} r \cos \theta = \cos \theta \frac{\partial}{\partial x} r = \cos \theta\frac{\partial r}{\partial x}.$$ Crucially, we are viewing $x$ and $r = r(x, y)$ as functions of $x$ and $y$ with $\theta$ constant. QED.