Let $z=f(x,y)$ where $x=r^2+s^2, y=2rs$, find $\partial^2z/\partial r^2$.
We first find $$\frac{\partial z}{\partial r}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial r}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial r}=2r\frac{\partial z}{\partial x}+2s\frac{\partial z}{\partial y}$$ then $$\begin{align}\frac{\partial^2 z}{\partial r^2} &=\frac{\partial}{\partial r} \left(2r\frac{\partial z}{\partial x}+2s\frac{\partial z}{\partial y}\right) \\ &=2\frac{\partial z}{\partial x}+2r\frac{\partial}{\partial r}\left(\frac{\partial z}{\partial x}\right)+2s\frac{\partial}{\partial r}\left(\frac{\partial z}{\partial y}\right) \\ \end{align} $$ put $$\begin{align}\frac{\partial}{\partial r}\left(\frac{\partial z}{\partial x}\right) &=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial x}\right)\frac{\partial x}{\partial r}+ \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial x}\right)\frac{\partial y}{\partial r}\\ \frac{\partial}{\partial r}\left(\frac{\partial z}{\partial y}\right) &=\frac{\partial}{\partial x}\left(\frac{\partial z}{\partial y}\right)\frac{\partial x}{\partial r}+ \frac{\partial}{\partial y}\left(\frac{\partial z}{\partial y}\right)\frac{\partial y}{\partial r} \end{align}$$ into $\partial^2z/\partial r^2$ we get $$\begin{align}\frac{\partial^2 z}{\partial r^2} &=2\frac{\partial z}{\partial x} + 4r^2\frac{\partial^2 z}{\partial r^2} + 8rs \frac{\partial^2x}{\partial x\partial y}+4s^2\frac{\partial^2 z}{\partial y^2} \end{align} $$ which is the correct answer.
However, if I try in this way: $$\begin{align}\frac{\partial^2 z}{\partial r^2} &=\frac{\partial}{\partial r} \left(2r\frac{\partial z}{\partial x}+2s\frac{\partial z}{\partial y}\right) \\ &=\frac{\partial}{\partial x} \left(2r\frac{\partial z}{\partial x}+2s\frac{\partial z}{\partial y}\right) \frac{\partial x}{\partial r} +\frac{\partial}{\partial y} \left(2r\frac{\partial z}{\partial x}+2s\frac{\partial z}{\partial y}\right) \frac{\partial y}{\partial r} \\ &=(2r)\left(2r \frac{\partial^2x}{\partial x^2}+2\frac{\partial z}{\partial x}\frac{\partial r}{\partial x}+2s\frac{\partial^2z}{\partial x\partial y}+2\frac{\partial z}{\partial y}\frac{\partial s}{\partial x}\right)+(2s)\left(2r\frac{\partial^2 z}{\partial x \partial y}+2\frac{\partial z}{\partial x}\frac{\partial r}{\partial y}+2s\frac{\partial^2 z}{\partial y^2}+2\frac{\partial z}{\partial y}\frac{\partial s}{\partial y}\right) \end{align} $$ It gives a different result $$4r^2\frac{\partial^2 z}{\partial x^2}+4\frac{\partial z}{\partial x}+8rs\frac{\partial^2 z}{\partial x\partial y}+\frac{2r}{s}\frac{\partial z}{\partial y}+\frac{2s}{r}\frac{\partial z}{\partial y}$$ Am I doing anything wrong in this way?
Additional question
If$$\frac{\partial z}{\partial r}=2r\frac{\partial z}{\partial x}+2s\frac{\partial z}{\partial y},$$ can$\cfrac{\partial}{\partial r}$written as$$\frac{\partial }{\partial r}=2r\frac{\partial }{\partial x}+2s\frac{\partial }{\partial y}$$ and so $$\begin{align}\frac{\partial^2 z}{\partial r^2} &=\frac{\partial}{\partial r} \left(2r\frac{\partial z}{\partial x}+2s\frac{\partial z}{\partial y}\right) \\ &=\left(2r\frac{\partial }{\partial x}+2s\frac{\partial }{\partial y}\right)\left(2r\frac{\partial z}{\partial x}+2s\frac{\partial z}{\partial y}\right) ? \end{align}$$
If so, why is that? And how can I proceed $\partial^2z/\partial r^2$ next in this way?
Following your derivatives within the computation of \begin{align}\frac{\partial^2 z}{\partial r^2} &=\ldots \\&=(2r)\left(2r \frac{\partial^2x}{\partial x^2}+2\frac{\partial z}{\partial x}\frac{\partial r}{\partial x}+2s\frac{\partial^2z}{\partial x\partial y}+2\frac{\partial z}{\partial y}\frac{\partial s}{\partial x}\right)+(2s)\left(2r\frac{\partial^2 z}{\partial x \partial y}+2\frac{\partial z}{\partial x}\frac{\partial r}{\partial y}+2s\frac{\partial^2 z}{\partial y^2}+2\frac{\partial z}{\partial y}\frac{\partial s}{\partial y}\right) \end{align} It seems to me, in the evaluation of the terms in $\frac{\partial z}{\partial y}$ you have incorrectly asumed that \begin{align} \frac{\partial s}{\partial x} &= \frac{1}{\frac{\partial s}{\partial s}}, \\ \frac{\partial r}{\partial x} &= \frac{1}{\frac{\partial r}{\partial s}} \end{align} Which is whay, from $$\frac{\partial x}{\partial s} = 2s$$ you have coefficients of $\frac{2r}{s}$ in $\frac{\partial z}{\partial y}$