Checking multivariable function's differentiability at $(0,0)$ I have a function: $$f(x,y) = \begin{cases} (x^2+y^2)\sin (x^2+y^2)^{-1} \ \ \ &\text{, if } x^2+y^2 \neq 0 \\ 0 &\text{, if } x^2+y^2=0 \end{cases}$$
I calculated the partial derivatives:
$$f_x(x,y)=2x\left( \sin (x^2+y^2)^{-1}-\frac{(x^2+y^2)\cos (x^2+y^2)^{-1}}{(x^2+y^2)^2} \right)$$
$$f_y(x,y)=2y\left( \sin (x^2+y^2)^{-1}-\frac{(x^2+y^2)\cos (x^2+y^2)^{-1}}{(x^2+y^2)^2} \right)$$
I'm stuck at checking the function's differentiability at point $(0,0)$, I use following definition :

$f$ is differentiable at point $(a,b)$ and $M,N\in \mathbb{R}$ $\leftrightarrow$
  $\lim_{h_1,h_2\to 0} \frac{f(a+h_1,b+h_2)-f(a,b)-Mh_2-Nh_2}{\sqrt{h_1^2+h_2^2}}=0$

Here's what I've done so far:
$$\lim_{h_1,h_2\to 0} \frac{(h_1^2+h_2^2)\sin(h_1^2+h_2^2)^{-1}-Mh^1-Nh_2}{\sqrt{h_1^2+h_2^2}}=$$ 
$$= \lim_{h_1,h_2\to 0} \frac{\sqrt{h_1^2+h_2^2}\sqrt{h_1^2+h_2^2}\sin(h_1^2+h_2^2)^{-1}-Mh^1-Nh_2}{\sqrt{h_1^2+h_2^2}}=$$
$$=\lim_{h_1,h_2\to 0} \frac{\sqrt{h_1^2+h_2^2}\left(\sqrt{h_1^2+h_2^2}\sin(h_1^2+h_2^2)^{-1}-\frac{Mh_1}{\sqrt{h_1^2+h_2^2}}-\frac{Nh_2}{\sqrt{h_1^2+h_2^2}}\right)}{\sqrt{h_1^2+h_2^2}}=$$
$$=\lim_{h_1,h_2\to 0} \sqrt{h_1^2+h_2^2}\sin(h_1^2+h_2^2)^{-1}-\frac{Mh_1-Nh_2}{\sqrt{h_1^2+h_2^2}}=$$

If a radial function $f(x,y) = g(x^2+y^2)$ is differentiable, its differential should be zero, by simmetry.
So you should check whether
$$
\lim_{(x,y)\to(0,0)}\frac{\sqrt{x^2+y^2}}{\sin(x^2+y^2)} = 0.
$$
But this is false even for $y=0$ where you get
$$
\lim_{x\to 0}\frac{|x|}{\sin(x^2)} = +\infty.
$$
So your function is not differentiable.
