Show that function is partially differentiable I have the following function:
$$F: \mathbb{R}^2 \rightarrow \mathbb{R}, ~~ (x,y) \rightarrow xy\frac{x^2-y^2}{x^2+y^2}$$
for $(x,y) \ne 0$ and $F(0,0) = 0$. I want to show that $F$ is partially differentiable twice.

So far I've figured out that for the function to be differentiable I have to show the differential quotient exists for both - $x$ and $y$. So starting with $x$ I'd treat $y$ as a constant and check for the limit:
$$\lim_{h \rightarrow 0} \frac{F(x_0+h) -F(x_0)}{h} = y\frac{(x_0+h)}{h}\frac{(x_0+h)^2-y^2}{(x_0+h)^2+y^2} - \frac{x_0y}{h}\frac{x_0^2-y^2}{x_0^2+y^2}$$
And unfortunately that is where I am stuck at the moment - I don't know how to get rid of the $h$ after expanding - maybe I've done something wrong?  
Thanks for helping out.
FunkyPeanut
 A: 
for the function to be differentiable I have to show the differential quotient exists for both

Yes. But in doing so, you should not forget single-variable calculus. Your strategy should be:


*

*Find $\dfrac{\partial F}{\partial x}$ and $\dfrac{\partial F}{\partial y}$ at points other than $(0,0)$. For this you use the standard machinery of Calculus I: product rule, quotient rule, etc. Since $F$ is a rational function, the partial  derivatives are known to exist.   

*Show that $\dfrac{\partial F}{\partial x}$ and $\dfrac{\partial F}{\partial y}$ also exist at $(0,0)$. This is where you need to fall back on the difference quotient definition, since $(0,0)$ is a special case of definition. Luckily, the difference quotients are simply zero all the way. So, both partials are equal to $0$ at $(0,0)$. 

*Optionally, find $\dfrac{\partial^2 F}{\partial x^2}$, $\dfrac{\partial^2 F}{\partial y^2}$ and $\dfrac{\partial^2 F}{\partial x\partial y}$  at points other than $(0,0)$. I don't think you are actually asked to do this. To justify that these derivatives exist, it suffices to appeal to the fact already mentioned: rational functions are differentiable in their domain.

*Show that $\dfrac{\partial^2 F}{\partial x^2}$, $\dfrac{\partial^2 F}{\partial y^2}$ and $\dfrac{\partial^2 F}{\partial x\partial y}$ also exist at $(0,0)$. This is like step 2, but you form difference quotients not from $F$, but from its 1st order derivatives (which you know from steps 1 and 2).

