This is from Q2-24 from Spivak's calculus on manifolds, where the question is to show that $D_{1,2}\ne D_{2,1}$, but it seems to me that instead it is showing that the second derivative is different from different directions. The question (and my answer) is as follows.
Consider the function defined as \begin{equation*} f(x)=\begin{cases} xy\frac{x^2-y^2}{x^2+y^2}\qquad&(x,y)\ne0\\ 0&(x,y)=0 \end{cases} \end{equation*} Taking the first derivatives and setting $x=0$ to the first case and $y=0$ in the second, \begin{equation*} \begin{split} D_yf(x,0)&=x\\ D_xf(0,y)&=-y \end{split} \end{equation*} Taking the second derivative \begin{equation*} \begin{split} D_{x,y}f(x,0)&=1\\ D_{y,x}f(0,y)&=-1 \end{split} \end{equation*} so as $(x,y)\rightarrow0$ the second derivatives are different.
However, I calculated the second derivative fully for both cases ($D_{x,y}$ and $D_{y,x}$), but it gives the same function. I do get the same result back when approaching from $x=0$ and $y=0$, but this seems to just show that the second derivative is direction dependent?
My result is $$D_{x,y}=D_{y,x}=\frac{x^2-y^2}{x^2+y^2}+8x^2y^2\left[\frac{x^2-y^2}{(x^2+y^2)^3}\right]$$ which I've also checked with symbolab which seems to agree with me.
