Show $ f(x,y)= 2x\sin(\frac{1}{\sqrt{x^2 + y^2}}) - x\cos(\frac{1}{\sqrt{x^2 + y^2}}) \sqrt{x^2+y^2}^{-1}$ not continuous at $(0,0)$ $f(x,y) = 2x\sin(\frac{1}{\sqrt{x^2 + y^2}}) - \frac{x\cos(\frac{1}{\sqrt{x^2 + y^2}})}{ \sqrt{x^2+y^2}}
$
I'm a bit puzzled. The statement is obviously true if you plot the function.
Formal: $\quad f(a_n,b_n) \not\rightarrow (0,0)\quad for \quad (a_n,b_n) \rightarrow (0,0)$
I've tried something along the lines of $\sin(\frac{1}{a_n}) = \sin(\pi[4n+1]) = 1$, but this attempt remained fruitless.

Let $x =r \cos(\phi), y = r\sin(\phi)$ then you have
\begin{align*}
\lim_{(x,y)\to(0,0)}f(x,y) &= \lim_{(x,y)\to(0,0)}2x\sin(\frac{1}{\sqrt{x^2 + y^2}}) - \frac{x\cos(\frac{1}{\sqrt{x^2 + y^2}})}{ \sqrt{x^2+y^2}} \\ 
&= \lim_{r\to 0}2 r\cos(\phi) \sin\left(\dfrac{1}{r}\right)- \dfrac{r\cos(\phi)\cos\left(\dfrac{1}{r}\right)}{r}
\end{align*}
Now $\lim_{r\to0}2r\cos(\phi)\sin\left(\frac{1}{r}\right)$ converges to $0$ indipendently from the angle $\phi$ but $\lim_{r\to 0} \cos(\phi)\cos\left(\frac{1}{r}\right)$ goes to $0$ dependently on the angle $\phi$ hence your function is not continuous in $(0,0)$
