Answer
(a) $\dfrac{\partial z }{\partial r}=(\cos \theta) \dfrac{\partial z}{\partial x} +(\sin \theta) \dfrac{\partial z}{\partial y}$
(b)$\dfrac{\partial z }{\partial \theta}=r \cos \theta (\dfrac{\partial z}{\partial y})-r \sin \theta \dfrac{\partial z}{\partial x} $
(c) $\dfrac{\partial^2 z }{\partial r \partial \theta}=-\sin \theta (\dfrac{\partial z}{\partial x})+ (\cos \theta) (\dfrac{\partial z}{\partial y}) -r \sin \theta \cos \theta (\dfrac{\partial^2 z}{\partial x^2})+r \sin \theta \cos \theta (\dfrac{\partial^2 z}{\partial y^2})+r( \cos^2 \theta -sin^2 \theta ) (\dfrac{\partial^2 z}{\partial y \partial x})$
Work Step by Step
a) $\dfrac{\partial z }{\partial r}=( \dfrac{\partial z}{\partial x}) \times (\dfrac{dx}{dr}) +( \dfrac{\partial z}{\partial y}) \times (\dfrac{dy}{dr})$
or, $\dfrac{\partial z }{\partial r}=(\cos \theta) \dfrac{\partial z}{\partial x} +(\sin \theta) \dfrac{\partial z}{\partial y}$
(b) $\dfrac{\partial z }{\partial \theta}=( \dfrac{\partial z}{\partial x}) \times (\dfrac{dx}{d\theta}) +( \dfrac{\partial z}{\partial y}) \times (\dfrac{dy}{d\theta})$
$\dfrac{\partial z }{\partial \theta}=(r \cos \theta) \times (\dfrac{\partial z}{\partial y})-(r \sin \theta) (\dfrac{\partial z}{\partial x})$
(c) $\dfrac{\partial z }{\partial r}=( \dfrac{\partial z}{\partial x}) \times (\dfrac{dx}{dr}) +( \dfrac{\partial z}{\partial y}) \times (\dfrac{dy}{dr})$
$\implies \dfrac{\partial z }{\partial r}=( \dfrac{\partial z}{\partial x}) \times (\cos \theta) +( \dfrac{\partial z}{\partial y}) \times (\sin \theta)$
and $\dfrac{\partial^2 z }{\partial r \partial \theta}=-\sin \theta (\dfrac{\partial z}{\partial x})+ (\cos \theta) (\dfrac{\partial z}{\partial y}) -r \sin \theta \cos \theta (\dfrac{\partial^2 z}{\partial x^2})+r \sin \theta \cos \theta (\dfrac{\partial^2 z}{\partial y^2})+r( \cos^2 \theta -\sin^2 \theta ) (\dfrac{\partial^2 z}{\partial y \partial x})$
