Answer
(a) $\dfrac{\partial z}{\partial r}=\cos \theta (\dfrac{\partial z}{\partial x}) +\sin \theta (\dfrac{\partial z}{\partial y}) ; \\
\dfrac{\partial z}{\partial \theta}=-r \ \sin \theta \dfrac{\partial z}{\partial x} +r \ \cos \theta \dfrac{\partial z}{\partial y} $
(b) $(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2= (\dfrac{\partial z}{\partial r})^2+\dfrac{1}{r^2}(\dfrac{\partial z}{\partial \theta})^2$
Work Step by Step
(a) $\dfrac{\partial z}{\partial r}=\cos \theta (\dfrac{\partial z}{\partial x}) +\sin \theta (\dfrac{\partial z}{\partial y}) ; \\
\dfrac{\partial z}{\partial \theta}=-r \ \sin \theta \dfrac{\partial z}{\partial x} +r \ \cos \theta \dfrac{\partial z}{\partial y} $
(b) $\dfrac{1}{r^2}(\dfrac{\partial z}{\partial \theta})^2=- (\dfrac{\partial z}{\partial r})^2 \\=-[(\dfrac{\partial z}{\partial x}) \cos \theta +(\dfrac{\partial z}{\partial y}) \sin \theta]^2$
$\implies -\dfrac{\partial z}{\partial x} \sin \theta +\dfrac{\partial z}{\partial y} \cos \theta^2=(\dfrac{\partial z}{\partial x} \sin \theta)^2 -2 (\dfrac{\partial z}{\partial x}) (\dfrac{\partial z}{\partial y}) \sin \theta \cos \theta+[\dfrac{\partial z}{\partial y} \cos \theta]^2$
$\implies (\dfrac{\partial z}{\partial x})^2 ( \cos^2 \theta +\sin^2 \theta)+(\dfrac{\partial z}{\partial y})^2 ( \cos^2 \theta +\sin^2 \theta)=(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2$
Since, $\cos^2 \theta +\sin^2 \theta=1$
Therefore, we have: $(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2= (\dfrac{\partial z}{\partial r})^2+\dfrac{1}{r^2}(\dfrac{\partial z}{\partial \theta})^2$ (proved)
