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) $\dfrac{\partial z}{\partial r}=(\dfrac{\partial z}{\partial x}) (\dfrac{\partial x}{\partial r})+(\dfrac{\partial z}{\partial y}) (\dfrac{\partial y}{\partial r})$
and $\dfrac{\partial z}{\partial r}==(\dfrac{\partial z}{\partial x}) \cos \theta +(\dfrac{\partial z}{\partial y}) \sin \theta \\\dfrac{\partial z}{\partial \theta}=(-r) \times \dfrac{\partial z}{\partial x} \sin \theta +r \times \dfrac{\partial z}{\partial y} \cos \theta$
(b) Here, we have $\dfrac{1}{r^2}(\dfrac{\partial z}{\partial \theta})^2=- (\dfrac{\partial z}{\partial r})^2=-[(\dfrac{\partial z}{\partial x}) \times \cos \theta +(\dfrac{\partial z}{\partial y}) \times \sin (\theta)]^2$
$\implies -\dfrac{\partial z}{\partial x} \times \sin \theta +(\dfrac{\partial z}{\partial y})^2 \times \cos^2 \theta=(\dfrac{\partial z}{\partial x} \sin \theta)^2 -2 (\dfrac{\partial z}{\partial x}) (\dfrac{\partial z}{\partial y}) \times \sin \theta \cos \theta+[\dfrac{\partial z}{\partial y}]^2 \times \cos ^2\theta$
or, $(\dfrac{\partial z}{\partial x})^2[ \cos^2 \theta +\sin^2 \theta]+(\dfrac{\partial z}{\partial y})^2 [ \cos^2 \theta +\sin^2 \theta]=(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2$
and $(\dfrac{\partial z}{\partial x})^2\times (1)+(\dfrac{\partial z}{\partial y})^2 \times (1)=(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2$
Thus, $(\dfrac{\partial z}{\partial x})^2+(\dfrac{\partial z}{\partial y})^2= (\dfrac{\partial z}{\partial r})^2+\dfrac{1}{r^2}(\dfrac{\partial z}{\partial \theta})^2$ (Verified)
