Answer
a) $\dfrac{\partial z }{\partial r}=( \dfrac{\partial z}{\partial x}) (\cos \theta) +( \dfrac{\partial z}{\partial y}) (\sin \theta)$
b) $\dfrac{\partial z }{\partial \theta}=(r \cos \theta) (\dfrac{\partial z}{\partial y})-(r \sin \theta) (\dfrac{\partial z}{\partial x})$
Work Step by Step
a) Here, we have $\dfrac{\partial z }{\partial r}=( \dfrac{\partial z}{\partial x}) (\dfrac{dx}{dr}) +( \dfrac{\partial z}{\partial y}) (\dfrac{dy}{dr})$
Plug in the given data, then we have $\dfrac{\partial z }{\partial r}=( \dfrac{\partial z}{\partial x}) (\cos \theta) +( \dfrac{\partial z}{\partial y}) (\sin \theta)$
b) Here, we have $\dfrac{\partial z }{\partial \theta}=( \dfrac{\partial z}{\partial x}) (\dfrac{dx}{d\theta}) +( \dfrac{\partial z}{\partial y}) (\dfrac{dy}{d\theta})$
Plug in the given data, then we have $\dfrac{\partial z }{\partial \theta}=(r \cos \theta) (\dfrac{\partial z}{\partial y})-(r \sin \theta) (\dfrac{\partial z}{\partial x})$
Hence, we have:
a) $\dfrac{\partial z }{\partial r}=( \dfrac{\partial z}{\partial x}) (\cos \theta) +( \dfrac{\partial z}{\partial y}) (\sin \theta)$
b) $\dfrac{\partial z }{\partial \theta}=(r \cos \theta) (\dfrac{\partial z}{\partial y})-(r \sin \theta) (\dfrac{\partial z}{\partial x})$