Find the value of $f_{xy}$ at the point (0,0) Let f be the function defined for all (x,y) as follows:
$f(x,y)=
\begin{cases}
\frac{xy(x^2-y^2)}{x^2+y^2}, &\text{if }(x,y)\ne(0,0)\\
0, &\text{if }(x,y)=(0,0)
\end{cases}$
What is the value of $\frac{\partial^2(f)}{\partial x\partial y}$ at $(0,0)$?
My work so far has been,
if we do the normal calculation, $\partial f/\partial x= \dfrac{x^4y-y^5+4x^2*y^3}{(x^2+y^2)^2}$
and then differentiate the result with respect to y, and we got,
$\dfrac{x^6+13x^4*y^2+27x^2*y^4-9y^6}{(x^2+y^2)^3}$
but I don't know how to proceed.
Before going this way, I am thinking of polar coordinates: $x=r\cos a$, $y=r\sin a$, then
$f(x,y) = f(r\cos a, r\sin a) = 1/2  \sin(4a)*r^2$
and $\frac{\partial f}{\partial x} = \frac{\partial f}{\partial r} \cdot \frac{\partial r}{\partial x} = 4r\sin(a)\cos(2a) = g$, and then differential this result with respect to y = rsin(a), I got $\partial g/\partial y = \partial g/\partial r \cdot \partial r/\partial y=4\cos(2a)$, so when when r, a goes to 0, the result should be 4? The solution says -1. Any ideas?
 A: Back to the definitions! Let us abbreviate $\partial/\partial z$ by $\partial_z$ and $\partial^2/(\partial z\partial t)$ by $\partial^2_{zt}$, then $\partial^2_{xy}f=\partial_xg$ where $g=\partial_yf$, that is, $$\partial^2_{xy}f(0,0)=\partial_xg(0,0)=\lim_{x\to0}\frac{g(x,0)-g(0,0)}x.$$
Likewise, for every $x$, $$g(x,0)=\partial_yf(x,0)=\lim_{y\to0}\frac{f(x,y)-f(x,0)}y.$$
Now the task is to check that the limits in the RHS above exist and to identify them. First, $f(x,0)=0$ for every $x$ hence, for every $y\ne0$, the ratio in the second limit is $$\frac{f(x,y)}y=\frac{x(x^2-y^2)}{x^2+y^2},$$ with limit when $y\to0$, for every $x$, $$g(x,0)=x.$$ Thus, the second derivative exists and is $$\partial^2_{xy}f(0,0)=+1.$$
Exercise: Apply this to the second derivative $\partial^2_{yx}f$ to show that $\partial^2_{yx}f(0,0)$ exists and $$\partial^2_{yx}f(0,0)=-1.$$
A: For $(x,y)\ne(0,0)$ you have obtained
$$f_x(x,y)={y(x^4+4x^2y^2-y^4)\over(x^2+y^2)^2},\quad f_y(x,y)={x(x^4-4x^2y^2-y^4)\over(x^2+y^2)^2}\ .\tag{1}$$
In order to compute the partial derivatives at $(0,0)$ we have to take recurse to the definition:
$$f_x(0,0)=\lim_{x\to0}{f(x,0)-f(0,0)\over x}=0,\quad f_y(0,0)=\lim_{y\to0}{f(0,y)-f(0,0)\over y}=0\ .$$
Now we can go at the mixed derivatives at $(0,0)$; again using the definition:
$$f_{xy}(0,0)=\lim_{y\to0}{f_x(0,y)-f_x(0,0)\over y}=\lim_{y\to0}{-y^5/y^4\over y}=-1\ ,$$
and in a similar way we obtain
$$f_{yx}(0,0)=\lim_{x\to0}{f_y(x,0)-f_y(0,0)\over x}=\lim_{x\to0}{x^5/x^4\over x}=1\ .$$
Note that the two mixed derivatives are unequal at $(0,0)$; the reason being that they are not continuous there. Formal derivation of $(1)$ gives
$$f_{xy}(x,y)={x^6 + 9 x^4 y^2 - 9 x^2 y^4 - y^6\over(x^2 + y^2)^3}\qquad\bigl((x,y)\ne(0,0)\bigr)\ ,$$
and here the right hand side has no limit when $(x,y)\to(0,0)$.
PS concerning polar coordinates: In polar coordinates the given function appears as
$$\tilde f(r,\phi)={1\over4}r^2 \>\sin(4\phi)\ .$$
Then
$${\partial f\over\partial x}={\tilde f\!}_r{\partial r\over\partial x}+{\tilde f\!}_\phi{\partial\phi\over\partial x}=\ldots$$
(you forgot the second term). Then you have to express $r_x$ and $\phi_x$ in terms of $r$ and $\phi$:
$${\partial r\over\partial x}={x\over r}=\cos\phi, \qquad{\partial\phi\over\partial x}={\partial\over\partial x}\arctan{y\over x}=\ldots=-{\sin\phi\over r}\ ;$$
and so on.
