Tell about derivatives $f(x,y)$. I have $f(x,y)=e^{-(x^2+y^2)^{-1}}$. Also $f(0,0)=0$. What I can tell about differentiable?

Of course I know $\frac{\partial f}{\partial x}$ and $\frac{\partial f}{\partial y}$. 

E.g. $$\frac{\partial f}{\partial x}=\frac{2x\cdot e^{-\frac{1}{x^2+y^2}}}{(-x^2-y^2)^2}$$

What I can tell about $\frac{\partial f}{\partial x}$ near $(0,0)$?

Thanks.
We have that

$$f_v(0,0)=\lim_{(h,k)\to(0,0)}\frac{e^{-(h^2+k^2)^{-1}}}{\sqrt{h^2+k^2}}=\lim_{u\to0^+}\frac{e^{-\frac1u}}{\sqrt{u}}=\lim_{v\to\infty}\frac  {\sqrt{v}}{e^{v}}= 0$$

then $f_x=f_y=0$.

Moreover since as $(x,y)\to (0,0)$

$$\frac{\partial f}{\partial x}=\frac{2x\cdot e^{-\frac{1}{x^2+y^2}}}{(x^2+y^2)^2}\to 0 \quad \frac{\partial f}{\partial y}=\frac{2y\cdot e^{-\frac{1}{x^2+y^2}}}{(x^2+y^2)^2}\to 0$$

by "Differentiability theorem" since all the partial derivatives exist and are continuous in a neighborhood of the point then (i.e. sufficient condition) the function is differentiable at that point.
The function $\phi (t)=e^{-1/|t|}$ for $t \neq 0$, $\phi (0)=0$ is an infinitely differentiable function from $\mathbb R$ into itself. Since  $f(x,y)=\phi (x^{2}+y^{2})$ it follows that $f$ is differentiable.