Answer
$\dfrac{\partial^2 z}{\partial r\partial s}=2y\dfrac{\partial^2 z}{\partial x^2}+2x \dfrac{\partial^2 z}{\partial x \partial y}+2y \dfrac{\partial^2 z}{\partial y^2}+2 \dfrac{\partial z}{\partial y}$
Work Step by Step
$\dfrac{\partial }{\partial r}(\dfrac{\partial z}{\partial x})=\dfrac{\partial^2 z}{\partial x^2} \times (2r) +\dfrac{\partial^2 z}{\partial y \partial x} \times (2s)$
and $\dfrac{\partial }{\partial r}(\dfrac{\partial z}{\partial y})=\dfrac{\partial^2 z}{\partial x\partial y} \times (2r) +\dfrac{\partial^2 z}{\partial y^2} \times (2s) $
$2s \times [\dfrac{\partial^2 z}{\partial x^2}(2r) +\dfrac{\partial^2 z}{\partial y \partial x}(2s)]+[\dfrac{\partial^2 z}{\partial x\partial y}(2r) +\dfrac{\partial^2 z}{\partial y^2}(2s)] +2 \times \dfrac{\partial z}{\partial y}=\dfrac{\partial^2 z}{\partial r\partial s}$
This can be re-written as: $\dfrac{\partial^2 z}{\partial x^2} \times (4rs) +(4r^2+4s^2) \times \dfrac{\partial^2 z}{\partial y \partial x}+ \dfrac{\partial^2 z}{\partial y^2}(4rs) +(2) \dfrac{\partial z}{\partial y}=\dfrac{\partial^2 z}{\partial r\partial s}$
We need to plug in $r^2+s^2=x$ and $2rs=y$ into the above equation.
Thus, we get
$\dfrac{\partial^2 z}{\partial r\partial s}=2y\dfrac{\partial^2 z}{\partial x^2}+2x \dfrac{\partial^2 z}{\partial x \partial y}+2y \dfrac{\partial^2 z}{\partial y^2}+2 \dfrac{\partial z}{\partial y}$ (VERIFIED)
