Answer
The answer is below.
Work Step by Step
(a) Let us have a function $f(x, y)=z$ which is expressed in polar form: $x=$ $r \cos \theta, r \sin \theta=y$
\[
\begin{aligned}
\frac{\partial z}{\partial r} &=\frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial r}+\frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial r} \\
&=\frac{\partial z}{\partial x} \cdot \frac{\partial}{\partial r}(r \cos \theta)+\frac{\partial z}{\partial y} \cdot \frac{\partial}{\partial r}(r \sin \theta) \\
&=\frac{\partial z}{\partial x} \cdot(\cos \theta)+\frac{\partial z}{\partial y} \cdot(\sin \theta) \\
r \frac{\partial z}{\partial r} &=r \cos \theta \frac{\partial z}{\partial x}+r \sin \theta \frac{\partial z}{\partial y}
\end{aligned}
\]
\[
=x \frac{\partial z}{\partial x}+y \frac{\partial z}{\partial y}
\]
(b) And:
$\frac{\partial z}{\partial \theta}=\frac{\partial z}{\partial x} \cdot \frac{\partial x}{\partial \theta}+\frac{\partial z}{\partial y} \cdot \frac{\partial y}{\partial \theta}$
$=\frac{\partial z}{\partial x} \cdot \frac{\partial}{\partial \theta}(r \cos \theta)+\frac{\partial z}{\partial y} \cdot \frac{\partial}{\partial \theta}(r \sin \theta)$
$=\frac{\partial z}{\partial x} \cdot(-r \sin \theta)+\frac{\partial z}{\partial y} \cdot(r \cos \theta)$
$\frac{\partial z}{\partial \theta}=-r \sin \theta \frac{\partial z}{\partial x}+r \cos \theta \frac{\partial z}{\partial y}$
$=x \frac{\partial z}{\partial y} -y \frac{\partial z}{\partial x}$
