Limit of a 2-variable function Given problem:
$$
\lim_{x\to0,y\to0}(1+x^2y^2)^{-1/(x^2+y^2)} 
.$$
I tried to do it with assigning y to $y = kx$, but that didn't help me at all. Also one point, I can't use L'Hospital's rule.

I assume here that you mean $\lim_{(x,y) \to (0,0)}$ then 
$$
\begin{align}
\lim_{(x,y) \to (0,0)} (1+x^2y^2)^{-\frac{1}{x^2+y^2}} = 
 \lim_{(x,y) \to (0,0)} \exp\left( \ln \left( (1+x^2y^2)^{-\frac{1}{x^2 + y^2}} \right) \right) &= \\ \\\lim_{(x,y) \to (0,0)} \exp\left( -\frac{\ln  (1+x^2y^2)}{x^2 + y^2}  \right) &= \\ \exp\left(  -\lim_{(x,y) \to (0,0)} \frac{\ln  (1+x^2y^2)}{x^2 + y^2}  \right) \tag{1}
\end{align}
$$
Assuming the limit exists and by continuity of the exponential.
Now if we look at the function which is clearly non-negative and using the fact that $\ln(1+x) \leq x$ we have 
$$
 0 \leq \frac{\ln  (1+x^2y^2)}{x^2 + y^2} \leq \frac{x^2y^2}{x^2+y^2} \\
 \lim_{(x,y) \to (0,0)} 0 \leq  \lim_{(x,y) \to (0,0)}\frac{\ln  (1+x^2y^2)}{x^2 + y^2} \leq \lim_{(x,y) \to (0,0)} \frac{x^2y^2}{x^2+y^2} = \\
 \lim_{r\to 0}\frac{r^4 \cos(t)^2\sin(t)^2}{r^2} = 0
$$
Hence 
$$
\lim_{(x,y) \to (0,0)}\frac{\ln  (1+x^2y^2)}{x^2 + y^2}  = 0 \tag{2}
$$
$(1)$ and $(2)$ allow you to conclude.
