<s>[INST] How to prove that $\frac{1}{x^4+y^4} e^{-\frac{1}{x^2+y^2}}\to0$ when $(x,y)\to (0,0) $? 
How can I calculate the limit 
  $$
\lim_{(x,y)\to (0,0)} \frac{1}{x^4+y^4} e^{-\frac{1}{x^2+y^2}}
$$
  and show that it is zero?

When switching to polar coordinates, I get:
$$
\lim_{(x,y)\to (0,0)} \frac{1}{r^4 (\cos^4 \theta+\sin^4 \theta) } e^{-\frac{1}{r^2}}
$$
but I have no idea how to show the term $\frac{1}{r^4 (\cos^4 \theta+\sin^4 \theta) }$ is bounded.
 [/INST] You need to show that $\cos^4\theta+\sin^4\theta$ never gets too small.  A nice way to do this is to show
$$\begin{align}
\cos^4\theta+\sin^4\theta
&=(\cos^2\theta+\sin^2\theta)^2-2\cos^2\theta\sin^2\theta\\
&=1-{1\over2}(2\sin\theta\cos\theta)^2\\
&=1-{1\over2}\sin^22\theta\\
&\ge1-{1\over2}\\
&={1\over2}
\end{align}$$
</s>