Prove that function is not C1 in every neighbourhood of (0,0) $f(x,y)=(x^2+y^2)\sin{\frac{1}{\sqrt{x^2+y^2}}}$
Wolfram partial derivative:
$$\frac{\partial f}{\partial x}=\frac{-x\cos{\frac{1}{\sqrt{x^2 + y^2}}}}{\sqrt{x^2 + y^2}} + 2 x \sin{\frac{1}{\sqrt{x^2 + y^2}}}
$$
I figured out its partial derivatives are actually continuos by trying to find the derivatives from definition.
What is the correct way to approach this?
 A: You want to prove that $f$ isn't $C^1$ is any neighborhood of the origin. In other words, you want to prove that at least one of the partial derivatives isn't continuous there.
The partial derivative with respect to the first coordinate, $\partial _1f$, at $(0,0)$ can be found as follows:
$$
  \lim \limits_{t\to 0}\left(\dfrac{f(t,0)}{t}\right)
  =\lim \limits_{t\to 0}\left(\dfrac{t^2\sin\left(\dfrac{1}{|t|}\right)}{t}\right)
  =\lim \limits_{t\to 0}\left(t\sin\left(\dfrac{1}{|t|}\right)\right)
  =0.
$$
So, $\partial _1f(0,0)=0$ (or, if you prefer, $\dfrac{\partial f}{\partial x}(0,0)=0$).
Thus,
$$\partial _1f(x,y)=
\begin{cases}
\frac{-x\cos\left(\frac{1}{\sqrt{x^2 + y^2}}\right)}{\sqrt{x^2 + y^2}} + 2x \sin\left({\frac{1}{\sqrt{x^2 + y^2}}}\right), &\text{if }(x,y)\neq(0,0)\\
0, &\text{if }(x,y)=(0,0)
\end{cases}$$
Now you want to prove that $\partial_1f$ isn't continuous at $(0,0)$. To do this note that $$\lim \limits_{(x,y)\to (0,0)}\left(2x \sin\left({\frac{1}{\sqrt{x^2 + y^2}}}\right)\right)=0.$$
Therefore $\partial _1f$ is continuous at the origin if, and only if, $$\lim \limits_{(x,y)\to (0,0)}\left(\frac{-x\cos\left(\frac{1}{\sqrt{x^2 + y^2}}\right)}{\sqrt{x^2 + y^2}}\right)=0.$$
But this limit doesn't exist. I leave it to you to prove this, with a hint: consider the sublimits along the curves $t\mapsto (t, 0)$.
