Find $f_{xy}(0,0)$ and $f_{yx}(0,0)$ for the following function:
$$f(x,y) =
\begin{cases}
\frac{xy}{x^{2}+y^{2}} & (x,y) \ne (0,0) \\
0 & (x,y) = (0,0)
\end{cases}$$
My try:
$$f_x =
\begin{cases}
\frac{y(y^2-x^2)}{(x^2+y^2)^2} & (x,y) \ne (0,0) \\
0 & (x,y) = (0,0)
\end{cases}$$
$$f_y =
\begin{cases}
\frac{x(x^2-y^2)}{(x^2+y^2)^2} & (x,y) \ne (0,0) \\
0 & (x,y) = (0,0)
\end{cases}$$
$$\color{blue}{f_x(0,0)}=\lim_{\Delta x \to 0}\frac{f\left(\Delta x,0\right)-f\left(0,0\right)}{ \Delta x}=\lim_{\Delta x \to 0}\frac{0-0}{\Delta x}=0$$
$$f_{xy}(0,0)=\lim_{\Delta y \to 0} \frac{f_x\left(0,\Delta y\right)-\color{blue}{f_x(0,0)}}{ \Delta y}=\lim_{\Delta y \to 0}\frac{
\frac{1}{\Delta y}-0}{\Delta y}=0$$
$$\color{red}{f_y(0,0)}=\lim_{\Delta y \to 0}\frac{f\left(0,\Delta y\right)-f\left(0,0\right)}{ \Delta y}=\lim_{\Delta y \to 0} \frac{0-0}{\Delta y}=0$$
$$f_{yx}(0,0)=\lim_{\Delta x \to 0} \frac{f_y\left(\Delta x,0\right)-\color{red}{f_y(0,0)}}{ \Delta x}=\lim_{\Delta x \to 0}\frac{\frac{1}{\Delta x}-0}{\Delta x}=0$$
If you have an alternative solution please first check mine.
Best Answer
Your only mistake here is \begin{equation} f_{xy}(0,0) = \lim_{h\rightarrow0}\frac{\frac{1}{h}}{h} = \lim_{h\rightarrow0}\frac{1}{h^2} = +\infty, \end{equation} and similarly $f_{yx}(0,0) =+\infty$. That being said, I think some remarks are well in order.
The function you are dealing with is a classical example illustrating some of the subtleties that arise when passing from single-variable calculus to multi-variate calculus. Embarassingly enough, I missed a lot of them myself.
Specifically, by setting $f(0,0)=0$ all 'horizontal' and 'vertical' sections of $f$ are continuous. By this I mean that for all fixed $y\in\mathbb{R}$, the single-variable function $f(\,\cdot\,,y)$ is continuous, and so is $f(x,\,\cdot\,)$ for any fixed $x\in\mathbb{R}$. Nonetheless $f$ is not continuous in $(0,0)$ as a multi-variate function, as you can see by examining its value on the diagonals $y=x$ and $y=-x$: \begin{equation} f(x,x) = \frac{x^2}{2x^2} = \frac{1}{2} \quad \forall x\neq0, \qquad f(x,-x) = \frac{-x^2}{2x^2} = -\frac{1}{2} \quad \forall x\neq0. \end{equation}
This is important because continuity is a necessary condition for differentiability. Thus this is a function which is derivable in the origin, but not differentiable. In particular, the partial derivatives are not continuous in the origin as, for example, \begin{equation} f_x(x,0) = 0, \qquad f_x(0,y) = \frac{1}{4y} \quad \forall y\neq0, \end{equation} hence $f_x(x,0)\rightarrow0$ for $x\rightarrow0$ whereas $f_x(0,y)\rightarrow+\infty$ for $y\rightarrow0^+$.