Answer
a) $\dfrac{\partial z}{\partial r}=(\dfrac{\partial z}{\partial x}) \cos \theta +(\dfrac{\partial z}{\partial y}) \sin \theta$
and
$\dfrac{\partial z}{\partial \theta}=-r \dfrac{\partial z}{\partial x} \sin \theta +r \dfrac{\partial z}{\partial y} \cos \theta$
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) $\dfrac{\partial z}{\partial r}=(\dfrac{\partial z}{\partial x}) (\dfrac{\partial x}{\partial r})+(\dfrac{\partial z}{\partial y}) (\dfrac{\partial y}{\partial r})=(\dfrac{\partial z}{\partial x}) \cos \theta +(\dfrac{\partial z}{\partial y}) \sin \theta$
and
$\dfrac{\partial z}{\partial \theta}=-r \dfrac{\partial z}{\partial x} \sin \theta +r \dfrac{\partial z}{\partial y} \cos \theta$
b) Prove that $(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2= (\dfrac{\partial z}{\partial r})^2+\dfrac{1}{r^2}(\dfrac{\partial z}{\partial \theta})^2$
Consider $RHS$
$\dfrac{1}{r^2}(\dfrac{\partial z}{\partial \theta})^2=- (\dfrac{\partial z}{\partial r})^2=-[(\dfrac{\partial z}{\partial x}) \cos \theta +(\dfrac{\partial z}{\partial y}) \sin \theta]^2$
This gives: $-\dfrac{\partial z}{\partial x} \sin \theta +\dfrac{\partial z}{\partial y} \cos \theta^2=(\dfrac{\partial z}{\partial x} \sin \theta)^2 -2 (\dfrac{\partial z}{\partial x}) (\dfrac{\partial z}{\partial y}) \sin \theta \cos \theta+[\dfrac{\partial z}{\partial y} \cos \theta]^2$
and $(\dfrac{\partial z}{\partial x})^2[ \cos^2 \theta +\sin^2 \theta]+(\dfrac{\partial z}{\partial y})^2 [ \cos^2 \theta +\sin^2 \theta]=(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2$
Hence, we get $(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2= (\dfrac{\partial z}{\partial r})^2+\dfrac{1}{r^2}(\dfrac{\partial z}{\partial \theta})^2$
