The task is to prove that $f(x,y)=\frac{x^2-4xy+2y^2}{e^{|x|+|y|}}$ is bounded and to find its biggest and lowest values. The only method of solving such a problem, that I have studied, is using Weierstrass' theorem for multiple variables:
If $f(x,y)$ is differentiable inside a compact set $C$ (compact space = closed and bounded set in $\mathbb{R}^n$) and continuous on the boundary of the set, then $f$ reaches its biggest and lowest values inside $C$.
This is an approximate translation of the theorem. The problem however is this: there is no explicitly defined compact space for the function, as it is defined everywhere in $\mathbb{R}^2$. I tried evaluating $\frac{x^2-4xy+2y^2}{e^{|x|+|y|}}$ because I have the feeling it's bigger than zero but I couldn't find any boundaries for the expression. The only limit I proved is $$\lim\limits_{(x,y)\to(\infty, \infty)}{\frac{x^2-4xy+2y^2}{e^{|x|+|y|}}}=0$$ but that doesn't really help me in any way. If I manage to find a compact space for the function, I would be able to procede with the problem. If you know any other method of proving the problem, it also helps.
Here is a proof that $f$ is bounded:
Let $f(x,y)=\frac{x^2-4xy+y^2}{e^{|x|+|y|}}$. Since you know $\lim_{(x,y)\to(\infty,\infty)}f(x,y)=0$, there exists a $R>0$ such that for every $(x,y)\in\mathbb R^2$ of distance $\sqrt{x^2+y^2}>R$, we have $|f(x,y)|<1$.
Now, since $C=\{(x,y)\in\mathbb R^2:\sqrt{x^2+y^2}\le R\}$ is compact and $f$ is continuous on it, by Weierstrass's theorem it must be bounded. That is, for some $A>0$ we have $|f(x,y)|<A$ for all $(x,y)\in C$.
Now, $|f(x,y)|\le\max\{1,A\}$ for every $(x,y)\in\mathbb R^2$.