# Show that $f$ is everywhere differentiable and the partials commute

Take the function $$f(x,y) = \begin{cases}\frac{x^3y -xy^3}{x^2+y^2} & (x,y) \neq (0,0) \\ 0 & (x,y) = (0,0) \end{cases}.$$ Show that it is everywhere differentiable and that $D_{1,2}f(0,0)$ and $D_{2,1}f(0,0)$ exist but are not equal.

I'm not sure how to show that it is everywhere differentiable. I would take the partials and say that they're continuous everywhere but (0,0) so it's differentiable everywhere but (0,0), but how do I show that it is also differentiable at (0,0)?

For the second part (that the two exist) is also confusing me. I can take the partials, but I don't know how to evaluate them at (0,0) since they would be undefined there.

To show that $f$ is differentiable, it suffices to show that all of its partial derivatives exist and are continuous.

As you've noted, this isn't a big deal where points other than $(0,0)$ are concerned; for $(0,0)$, you'll just need to deal with the definition: $$f_x(0,0)=\lim_{h\to0}\frac{f(0+h,0)-f(0,0)}{h},\qquad f_y(0,0)=\lim_{h\to0}\frac{f(0,0+h)-f(0,0)}{h}.$$

Lastly, you need to show that the partials are continuous at $(0,0)$; this means taking the partial derivatives away from $(0,0)$, and showing that they converge to $f_x(0,0)$ and $f_y(0,0)$ (respectively) as $(x,y)\to(0,0)$.

For the second part: you'll now have expressions for the partial derivatives; now, the fun starts again, needing to compute MORE partial derivatives at the origin.

Let's calculate the partials:

$$\partial_xf(x,y)=y\frac{x^4+4x^2y^2-y^4}{(x^2+y^2)^2}\qquad\partial_yf(x,y)=x\frac{x^4-4x^2y^2-y^4}{(x^2+y^2)^2}$$

These go to zero as $(x,y)\to(0,0)$ (note that $|x^4+4x^2y^2-y^4|\le2|x^4+2x^2y^2+y^4|$ $=2(x^2+y^2)^2$). Also, the original function is bounded as

$$\left|\frac{x^3y-xy^3}{x^2+y^2}\right|\le\left|\frac{xy(x^2+y^2)}{x^2+y^2}\right|\le|xy|\le x^2+y^2,$$

so the partials exist at $(0,0)$ and are both zero. Thus the partials are continuous everywhere.

Returning to the calculation of the second-order partials: Plug in $x=0$ to $\partial_x f$ to get $\partial_x f|_{x=0}=-y$, and similarly $\partial_y f|_{y=0}=x$. Thus $\partial_y\partial_x f(0,0)$ exists and is $-1$, while $\partial_x\partial_y f$ exists and is $1$.