Show that both mixed partial derivatives exist at the origin but are not equal $$f(x,y) = \begin{cases} \displaystyle \frac{xy(x^2-y^2)}{x^2+y^2} & \text{if } (x,y) \neq (0,0), \\ 0 & \text{if } (x,y) = (0,0). \end{cases}$$
I tried finding both mixed partial derivatives but they ended up being the same for that function. I must be failing to take into account something dealing with the fact that it is piece-wise. I still need to show the mixed partial derivatives exist. How can I do all of this?
 A: Use the definition of partial derivative:
$$
  f_x(0,0)
~=~
  \lim_{h\to 0}
  \frac{f(h,0)-f(0,0)} h
~=~
  \lim_{h\to 0}
  \frac{\frac{h\cdot 0(h^2-0^2)}{h^2+0}-0} h
~=~
  \lim_{h\to 0}
  \frac{0}{h}
~=~
  0
$$
A similar computation shows that $f_y(0,0)=0$ too, so that $(0,0)$ is a critical point (i.e. $\nabla f(0,0)=\binom 00$).
EDIT
Now that we know the values of $f_x(0,0)$ and $f_y(0,0)$ we can compute $f_{xy}(0,0)$ and $f_{yx}(0,0)$:
$$
f_{xy}(0,0)
~=~
\lim_{k\to 0} \frac{f_x(0,k)-f_x(0,0)}k
~=~
\lim_{k\to 0} \frac{f_x(0,k)}k
$$
and
$$
f_{yx}(0,0)
~=~
\lim_{h\to 0} \frac{f_y(h,0)-f_y(0,0)}h
~=~
\lim_{h\to 0} \frac{f_y(h,0)}h
$$
First, note that for $(x,y)\neq(0,0)$ you have
$$
f_x(x,y)=\frac{y\big(x^4+4x^2y^2-y^4\big)}{\big(x^2+y^2\big)^2}
$$
and
$$
f_y(x,y)=\frac{x\big(x^4-4x^2y^2-y^4\big)}{\big(x^2+y^2\big)^2}
$$
so that for $h,k\neq 0$
$$
f_x(0,k)=-k
\quad\text{and}\quad
f_y(h,0)=h
$$
Putting all together:
$$
f_{xy}(0,0)
~=~
\lim_{k\to 0} \frac{f_x(0,k)}k
~=~
\lim_{k\to 0}\frac{-k}{k}
~=~
-1
$$
and
$$
f_{yx}(0,0)
~=~
\lim_{h\to 0} \frac{f_y(h,0)-f_y(0,0)}h
~=~
\lim_{h\to 0} \frac{f_y(h,0)}h
~=~
\lim_{h\to 0}\frac{h}{h}
~=~
1
$$
so
$$
f_{xy}(0,0)~=~-1 ~~\neq~~ 1~=~f_{yx}(0,0)
$$
