Limit in two variables of $x e^{-y^2/x}$ 
I'm trying to compute the limit
$$\lim_{(x,y)\rightarrow (0,0)} x e^{-y^2/x}$$

In order to compute it, I would just use the following bound:
$$e^{-\frac{-y^2}{x}} \leq 1$$ and hence
$$|x e^{-y^2/x}|$$ and hence the limit is $0$. Is it right?

Using polar coordinates I obtain
$\lim_{r \rightarrow 0} r cos(\theta) e^{- r \frac{sin^2(\theta)}{cos(\theta)}}$
why isn't this $0$?
 A: If $$f(x,y)=xe^{-\frac{y^2}{x}}$$
then
$$F(y)=f(y^3,y)=y^3e^{-\frac 1y}$$
and
$$\lim_{y\to 0^-}F(y)=-\infty$$
Your limit does not exist.
Observe also that
$$\lim_{y\to 0}f(y^2,y)=0$$
A: A limit $L$ of a multivariate function $f$ exists at a point $P$ if any only the limits taken on any path at $P$ is equal to $L$.
This is quite technical, so I'll simplify it. Take $P=(x_0, y_0)$ (In your case, $P=(0,0)$). Then, a path that meets $P$ is any continuous function $y(x)$ where $y(x_0)=y_0$. (I.e. $P$ is a point on the path) Likewise, we might also use $x(y)$ s.t. $x(y_0)=x_0$.
Then we may take: $$\lim_{(x,y)\to (x_0, y_0)}f(x,y)=\lim_{x\to x_0}f(x,y(x))$$
If the right-hand limit is the same no matter what $y(x)$ we pick (subject to the above constraint that $y(x_0)=y_0$), then it's limit $L$ is the limit of the left-hand side. If there is any function where they are unequal or the limit doesn't exist, the multivariate function has no limit.
hamam_Abdallah pointed out in his answer that taking $x(y)=y^3$, we have that $x(0)=0$ so the constraint is met, and then $$\lim_{(x,y)\to (0,0)}f(x,y)=\lim_{y\to 0}y^3e^{-\frac 1y}$$
which doesn't exist (The LHS limit is $-\infty$ and the RHS limit is $0$), and so the general limit doesn't exist.
