Partial derivative of $xy\frac{x^2-y^2}{x^2+y^2}$ I am asked to show, that

$f(x,y)=\begin{cases} xy\frac{x^2-y^2}{x^2+y^2}\space\text{for}\, (x,y)\neq (0,0)\\ 0\space\text{for}\, (x,y)=(0,0)\end{cases}$
is everywhere two times partial differentiable, but it is still $D_1D_2f(0,0)\neq D_2D_1 f(0,0)$

But this does not make much sense in my opinion. Since $f(0,0)=0$ hence it should be equal?
I calculated $\frac{\partial^2 f(x,y)}{\partial x\partial y}$ and $\frac{\partial^2 f(x,y)}{\partial y\partial x}$ and I got in both cases the same result (for $(x,y)\neq (0,0)$) which is:
$\frac{x^6+9x^4y^2-9x^2y^4-y^6}{(x^2+y^2)^3}$
Wolframalpha says, that this is correct.
So is there a mistake in the task? Or do I not understand it?
Also, when I want to show, that $f$ is everywhere two times partial differentiable. Is it enough to calculate $\frac{\partial^2 f}{\partial^2 x}$ and $\frac{\partial^2 f}{\partial^2 y}$ and not needed to go by the definition, since we know, that this function is differentiable as a function in $\mathbb{R}$?
What do you think?
Thanks in advance.
 A: The limit definition of partial derivative at $(x,y)=(0,0)$:
$$f_x(0,0)=\frac{\partial f}{\partial x}(0,0)=\lim_{h\to 0} \frac{f(0+h,0)-f(0,0)}{h}=\frac{\frac{(0+h)\cdot 0\cdot ((0+h)^2-0^2)}{(0+h)^2+0^2}-0}{h}=0;\\
f_y(0,0)=\frac{\partial f}{\partial y}(0,0)=\lim_{h\to 0} \frac{f(0,0+h)-f(0,0)}{h}=\frac{\frac{0\cdot(0+h)\cdot (0^2-(0+h)^2)}{0^2+(0+h)^2}-0}{h}=0.$$
Note that:
$$f_x(x,y)=\frac{\partial f}{\partial x}=\frac{\partial f}{\partial x}\left(\frac{x^3y-xy^3}{x^2+y^2}\right)=\frac{y(x^4+4x^2y^2-y^4)}{(x^2+y^2)^2};\\
f_y(x,y)=\frac{\partial f}{\partial y}=\frac{\partial f}{\partial y}\left(\frac{x^3y-xy^3}{x^2+y^2}\right)=\frac{x^5-4x^3y^2-xy^4}{(x^2+y^2)^2}.\\
$$
The partial derivative of partial derivative:
$$\frac{\partial^2 f}{\partial x\partial y}(0,0)=
\lim_{h\to 0} \frac{f_x(0,0+h)-f_x(0,0)}{h}=\\
\lim_{h\to 0} \frac{\frac{(0+h)(0^4+4\cdot 0^2(0+h)^2-(0+h)^4)}{(0^2+(0+h)^2)^2}-0}{h}=\lim_{h\to 0} \frac{-h^5}{h^5}=-1;\\
\frac{\partial^2 f}{\partial y\partial x}(0,0)=
\lim_{h\to 0} \frac{f_y(0+h,0)-f_y(0,0)}{h}=\\
\lim_{h\to 0} \frac{\frac{(0+h)^5-4(0+h)^3\cdot 0^2-(0+h)\cdot 0^4}{((0+h)^2+0^2)^2}-0}{h}=\lim_{h\to 0} \frac{h^5}{h^5}=1.$$
A: First a comment
You wrote But this does not make much sense in my opinion. Since $f(0,0)=0$...
How can you induce an affirmation on second derivatives of a map based on its value at a point?
Second regarding partial second derivatives
You have for $(x,y)\neq(0,0)$
$$\begin{cases} 
\frac{\partial f}{\partial x}(x,y) = \frac{(x^2+y^2)(3x^2y-y^3)+2x^2y(y^2-x^2)}{(x^2+y^2)^2}\\ 
\frac{\partial f}{\partial y}(x,y) = \frac{(x^2+y^2)(x^3-3x y^2)+2xy^2(y^2-x^2)}{(x^2+y^2)^2} 
\end{cases}$$
In particular
$$\begin{align*} 
\frac{\partial f}{\partial x}(0,y) &= -y \text{ for } y \neq 0\\ 
\frac{\partial f}{\partial x}(x,0) &= x \text{ for } x \neq 0 
\end{align*}$$
 which implies
$$
\frac{\partial^2 f}{\partial x \partial y}(0,0) =1 \neq -1 = \frac{\partial^2 f}{\partial y \partial x}(0,0)$$
You can have a look here for more details.
