$f(x,y) = e^\left({\frac{1}{x^2+y^2-1}}\right)$ continuity for $x^2+y^2<1$ we have:
$$f(x,y) = e^\left({\frac{1}{x^2+y^2-1}}\right)$$
for $x^2+y^2\ge 1$ we have:
$$f(x,y) = 0$$
I'm asked in which sets these functions are continuous. That's what I did:
for the set $\{x^2+y^2<1\}$, we have that the functions is continuous, as there is no breaks in the function. 
for the set $\{x^2+y^2=1\}$ we have the function undefined, because we get:
$$e^\left({\frac{1}{1-1}}\right)$$
for the set $\{x^2+y^2>1\}$
we have that the limit of $f(x,y)$ when $(x,y)$ 'goes to infinity' we have that $f(x,y)$ goes to $1$, which does not approaches $0$, therefore for this set, the function is not continuous. So I concluded that it is continuous only for the set $\{x^2+y^2<1\}$. Am I rigth?

Update:
All I'm trying to do is to analise the behaviour of the function
  outside the set $\{x^2+y^2<1\}$. That is,  I need to find the
  'directions' where, when I approach the boundaries, I approach $0$.

 A: Hint: On $\mathbb {R}$ define $g(t) = \exp (1/(t^2-1)), |t| < 1, g(t) = 0, |t|\ge 1.$ If you can show $g$ is continous on $\mathbb {R},$ then $f(x,y) = g(x^2+y^2)$ is the composition of continuous functions, hence is continuous.
A: Note that in your case function $f$ depends only on the expression $x^2 + y^2$. 
Therefore we can denote $t := x^2 + y^2 - 1$, so that your functions $f$ now depends only on one variable:
$$
f \left( x, y \right) = f \left(x^2 + y^2\right) = 
\begin{cases}
e^{\frac{1}{x^2+y^2-1}}, & x^2+y^2 < 1 \\
0, & x^2+y^2\geq 1
\end{cases}
\implies 
f\left(t\right) = 
\begin{cases}
e^{\,1/t}, & t < 0 \\
0, & t\geq 0
\end{cases}
$$
Now, recall that the function  $f$ is called continuous at a point $t_0$ if lefts and right limits of $f(t)$ as $x$ approaches $t_0$ through the domain of $f$ exist and are equal to $f\left(t_0\right)$:
$$
f\left(t\right)\rvert_{t_0}\in \mathcal C^0 \iff 
\lim_{t\to t_0^+} \ f\left( t\right) = 
\lim_{t\to t_0^-} \ f\left( t\right) = 
f\left( t_0\right).
$$
In our case $t_0 = 0$, so that 
$$
\begin{aligned}
f\left(t_0\right) & = f\left(0\right) = 0,
\\
f^+\left(t_0\right) & = \lim_{t\to 0^+} \ f\left( t\right) = 0,
\\
f^-\left(t_0\right) & = \lim_{t\to 0^-} \ f\left( t\right) = 
\lim_{t\to 0^-} e^{\,1/t} = e^{-\infty} =  0
\end{aligned}
$$
Thus, we conclude that $f$ is continuous.
