Answer
$\dfrac{\partial ^2z }{\partial x^2}+\dfrac{\partial ^2z }{\partial y^2}=\dfrac{\partial^2z}{\partial r^2}+\dfrac{1}{r^2} \dfrac{\partial^2z}{\partial \theta^2}+\dfrac{1}{r} \dfrac{\partial z}{\partial r} $
Work Step by Step
Here, we have $\dfrac{\partial z }{\partial x}=( \dfrac{\partial z}{\partial r}) ( \dfrac{\partial r}{\partial x})+( \dfrac{\partial z}{\partial \theta}) ( \dfrac{\partial \theta}{\partial x})=\cos \theta (\dfrac{\partial z}{\partial r})- \dfrac{\sin \theta}{r} (\dfrac{\partial z}{\partial \theta}) $
$\dfrac{\partial z }{\partial y}=( \dfrac{\partial z}{\partial r}) ( \dfrac{\partial r}{\partial y})+( \dfrac{\partial z}{\partial \theta}) ( \dfrac{\partial \theta}{\partial y})=\sin \theta (\dfrac{\partial z}{\partial r})+ \dfrac{\cos \theta}{r} (\dfrac{\partial z}{\partial \theta}) $
Now, $\dfrac{\partial ^2z }{\partial x^2}=\cos^2 \theta (\dfrac{\partial^2z}{\partial r^2})- \dfrac{2 \sin \theta \cos \theta }{r} (\dfrac{\partial^2 z}{\partial r\partial \theta})+\dfrac{\sin^2 \theta}{r^2} (\dfrac{\partial^2 z}{\partial \theta^2})+\dfrac{\sin^2 \theta}{r} (\dfrac{\partial z}{\partial \theta})+\dfrac{2 \sin \theta \cos \theta }{r^2} (\dfrac{\partial z}{\partial \theta}) $ ...(1)
$\dfrac{\partial ^2z }{\partial y^2}=\sin^2 \theta (\dfrac{\partial^2z}{\partial r^2})+ \dfrac{2 \sin \theta \cos \theta }{r} (\dfrac{\partial^2 z}{\partial r\partial \theta})+\dfrac{\cos^2 \theta}{r^2} (\dfrac{\partial^2 z}{\partial \theta^2})+\dfrac{\cos^2 \theta}{r} (\dfrac{\partial z}{\partial r})-\dfrac{2 \sin \theta \cos \theta }{r^2} (\dfrac{\partial z}{\partial \theta}) $ ...(2)
Using equations (1) and (2), we have
$\dfrac{\partial ^2z }{\partial x^2}+\dfrac{\partial ^2z }{\partial y^2}=\dfrac{\partial^2z}{\partial r^2}+\dfrac{1}{r^2} \dfrac{\partial^2z}{\partial \theta^2}+\dfrac{1}{r} \dfrac{\partial z}{\partial r} $
