Finding limit of $ \lim \limits_{x,y \to 0,0}{(1 + x^2 y^2)}^{-\frac{1}{x^2 + y^2}}$ Here is my limit:
$$ \lim \limits_{x,y \to 0,0}{(1 + x^2 y^2)}^{-\frac{1}{x^2 + y^2}}$$
I have learned two methods. One where we replace y with for example $y = kx $ (because $y = y_0 + k(x - x_0)$ and  $y_0 = 0, x_0 = 0$). Or with $x = r *cos(\phi)$ and $x = r *sin(\phi)$ where $r \to 0$.
Neither seem to help me at the moment (or at least when I tried solving with both I didn't get a good answer.
It kind of seems like I could use $ \lim \limits_{x \to \infty}{(1 + \frac{1}{x})}^{x} = e$, but I tried and also couldn't get a decent answer.
Any ideas?

We can use your idea of setting $x= r \cos \theta$, $y=r \sin \theta$ and the limit becomes 
$$\lim_{r \rightarrow 0} \left(1+\frac{r^4 \sin^2 2\theta}{4} \right)^{-\frac{1}{r^2}}$$
For a given $r$, the maximum and minimum values of this function are 
$1$ and $\left( 1+\frac{r^4}{4} \right)^{-\frac{1}{r^2}}$  obtained by setting $\theta =0$ and  $\theta = \frac{\pi}{4}$ respectively. 
The second limit as $r\rightarrow 0$ is 
$\lim_{r\rightarrow 0}\left( 1+\frac{r^4}{4} \right)^{-\frac{1}{r^2}} = \lim_{x\rightarrow \infty} \left( 1+\frac{1}{x^2} \right)^{\frac{x}{2}} = 1$ 
where we have taken $x = \frac{2}{r^2}$ 
Because this the minimum value the function can take on the circle, we can say the following:
For any $\epsilon >0$ there exists $r$ such that $x^2 + y^2 < r^2 \Rightarrow |(1+x^2y^2)^{-\frac{1}{x^2+y^2}} -1| < \epsilon$ so the limit is $1$. 
