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.

Using Polar coordinates, $x = r\cos\theta$ and $y = r\sin\theta$, the problem boils down to $$ \lim_{r\to0}\space(1 + r^4\cos^2\theta\sin^2\theta)^{\frac{-1}{r^2}}$$
Now prove that independent of the value of $\theta$, the limit exists and solve for it then.
EDIT:
Now let $$L = \lim_{r\to0}\space(1 + r^4\cos^2\theta\sin^2\theta)^{\frac{-1}{r^2}}$$
Taking $\log$ on both sides,
$$\log L = \lim_{r\to0}\space -\dfrac{\log(1 + r^4\cos^2\theta\sin^2\theta)}{r^2}$$As $\log (1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \cdots$,
$$\log L = \lim_{r\to0}\space - \dfrac{1}{r^2}\left (r^4\cos^2\theta\sin^2\theta - \dfrac{\left (r^4\cos^2\theta\sin^2\theta\right )^2}{2}) + \cdots \right ) = 0$$
$$\implies \log L =0 \implies L = 1$$
