Find all the points of continuity of $f :\mathbb R^2 \to \mathbb R$ 
Find all the points of continuity of the $ f : \mathbb R ^2 \to
 \mathbb R $, $$ f (x, y) = \begin{cases} | yx ^{-2} | \cdot e^{- | yx^{- 2} |},
x \neq 0 \\ 0, x = 0 \end {cases} $$

$f$ is continuity for $\mathbb R^2 \setminus Z$, where $Z=\{(x,y):x=0\}$
I need to check if $\lim_{(x,y)\to (0,y)} | yx ^{-2} | \cdot e^{- | yx^{- 2} |}=0$. I check it for some sequence for example $(0+\frac 1n , y+ \frac 1n)$ and I get $0$, but I can't prove it for every $(x_n,y_n) \to (0,0)$.
 A: We have that for $y_{1}>0$. Then in a small neighbourhood of $(0,y_{1})$ which excludes $(0,0)$ we have  $\displaystyle\bigg|(\frac{y}{x^{2}})e^{-\frac{y}{x^{2}}}\bigg|\leq \frac{x^{2}}{y}$ and this tends to $0$ as $(x,y)\to (0,y_{1})$ .
For the point $(0,0)$ consider the sequence $(\frac{1}{n},\frac{1}{n^{2}})$ . Then we have the limit as $e^{-1}$ . (Alternatively consider approaching the origin by the curves $y=mx^{2}$ for different values of $m$)
But when we consider the sequence $(\frac{1}{n},\frac{1}{n})$ we end up with the limit as $\displaystyle\lim_{n\to \infty}n\cdot e^{-n}$ which tends to $0$. Hence the limit does $\displaystyle\lim_{(x,y)\to (0,0)}f(x,y)$ does not exist as it is not unique.
And it is continuous when $x\neq 0$ because it is a composition of continuous functions . $\frac{y}{x^{2}}$ is continuous when $x\neq 0$ and $xe^{-x}$ is also continuous . So composing them we get that $\frac{y}{x^{2}}e^{-\frac{y}{x^{2}}}$ is continuous when $x\neq 0$.
If $y_{1}<0$ then we have in a small neighbourhood of $(0,y_{1})$ which is contained in the lower half plane , we have $\bigg|\frac{y}{x^{2}}e^{-\frac{y}{x^{2}}}\bigg|\geq \bigg|\frac{M}{x^{2}}e^{\frac{M}{x^{2}}}\bigg|$ for some $M> 0$.
But  $\displaystyle \frac{M}{x^{2}}e^{\frac{M}{x^{2}}}$ is unbounded when $x\to 0$ and hence the function is discontinuous.
So the function is continuous on $\Bbb{R}^{2}\setminus\{(0,y):y\in (-\infty,0]\}$
