Answer
$\dfrac{\partial^2 z}{\partial r\partial s}=\dfrac{\partial^2 z}{\partial x^2}(2y) +(2x) \dfrac{\partial^2 z}{\partial x \partial y}+(2y) \dfrac{\partial^2 z}{\partial y^2}+(2) \dfrac{\partial z}{\partial y}$
Work Step by Step
$\dfrac{\partial }{\partial r}(\dfrac{\partial z}{\partial x})=\dfrac{\partial^2 z}{\partial x^2}(2r) +\dfrac{\partial^2 z}{\partial y \partial x}(2s)$
and $\dfrac{\partial }{\partial r}(\dfrac{\partial z}{\partial y})=\dfrac{\partial^2 z}{\partial x\partial y}(2r) +\dfrac{\partial^2 z}{\partial y^2}(2s) $
$\dfrac{\partial^2 z}{\partial r\partial s}=2s [\dfrac{\partial^2 z}{\partial x^2}(2r) +\dfrac{\partial^2 z}{\partial y \partial x}(2s)]+(2r) [\dfrac{\partial^2 z}{\partial x\partial y}(2r) +\dfrac{\partial^2 z}{\partial y^2}(2s)] +) \dfrac{\partial z }{\partial y} (2)$
Now, $\dfrac{\partial^2 z}{\partial r\partial s}=\dfrac{\partial^2 z}{\partial x^2}(4rs) +\dfrac{\partial^2 z}{\partial y \partial x}(4r^2+4s^2)+ \dfrac{\partial^2 z}{\partial y^2}(4rs) +(2) \dfrac{\partial z}{\partial y}$
Substitute $r^2+s^2=x \\ 2rs=y$
$\dfrac{\partial^2 z}{\partial r\partial s}=(2y) \ \dfrac{\partial^2 z}{\partial x^2} +(2x) \ \dfrac{\partial^2 z}{\partial x \partial y}+(2y) \ \dfrac{\partial^2 z}{\partial y^2}+(2) \dfrac{\partial z}{\partial y}$
