differentiability of $f(x,y)=xy\sin\left({1\over x^2+y^2}\right)$ Let $f(x,y)=xy\sin\left({1\over x^2+y^2}\right)$ if $(x,y)\neq (0,0)$ and $0$ if $(x,y)=(0,0)$. Determine the points in which $f$ is differentiable
I know that $f(x,y)$ is differentiable at $(x_0,y_0)$ iff there exists a linear function $l:\mathbb R^2 \to \mathbb R$ so that:
$$\lim_{(x,y)\to (x_0,y_0)}{f(x,y)-f(x_0,y_0)-l(x-x_0,y-y_0)\over \|\vec x - \vec x_0\|}=0$$
If I want to see the points in which $f$ is differentiable then I need to pick an arbitry point $(x_0,y_0)$ but then the last expression becomes very complicated; Is there an easy way do it?
I would really appreciate your help :)
 A: The derivative exists at a point $(x_0,y_0)$ if the partial derivatives exist in an open neighborhood and are continuous at $(x_0,y_0)$.
This is clearly the case everywhere except possibly $(0,0)$ so the behavior at that point must be examined more closely.
Note that
$$f_x(x,y) = y\sin\left({1\over x^2+y^2}\right)- \frac{2x^2y}{(x^2+y^2)^2}\cos\left({1\over x^2+y^2}\right),\\ f_y(x,y) = x\sin\left({1\over x^2+y^2}\right)- \frac{2xy^2}{(x^2+y^2)^2}\cos\left({1\over x^2+y^2}\right).$$
Along a path with $y=0$
$$\lim_{(x,y) \rightarrow (0,0), y=0}f_x(x,y)=0$$
Along a path with $y=x$
$$\lim_{(x,y) \rightarrow (0,0), y=x}f_x(x,y)=\lim_{x \rightarrow 0}x\sin\left({1\over 2x^2}\right)-\lim_{x \rightarrow 0}\frac{2x^3}{4x^4}\cos\left({1\over 2x^2}\right)=0 \pm \infty.$$
Hence, the partial derivatives are not continuous at $(0,0)$.
Since continuity of the partial derivatives is a sufficient condition for the existence of the derivative, we cannot conclude non-existence at this point.
Notice that
$$\frac{|f(x,y)|}{\sqrt{x^2+y^2}} \leqslant \frac{|x||y|}{\sqrt{x^2+y^2}}\leqslant \min(|x|,|y|).$$
So try using $l:(x,y)\mapsto 0$.
Then 
$$\lim_{(x,y) \rightarrow (0,0)}\frac{f(x,y)-f(0,0)-l(x,y)}{\sqrt{x^2+y^2}}=\lim_{(x,y) \rightarrow (0,0)}\frac{xy}{\sqrt{x^2+y^2}}\sin\left({1\over x^2+y^2}\right)=0,$$
and the derivative at $(0,0)$ is the zero linear operator.
