$$A=\left\{\frac{m^4+2n^2}{2m^2-m^2n+n^2}:m,n \in\Bbb N\right\}$$
$\sup A$ doesn't exist, because $A$ is not bounded above:
$$\begin{align}n&=1 \\ \frac{m^4+2}{m^2+1} &\le M \\ \frac{m^4+2}{m^2} &\le M \\ m^2+\frac{2}{m^2} &\le M \\ m^2&\le M\end{align}$$
But I can choose $m=\lfloor M \rfloor+1$ from Archimedean property.
I think $\inf A=3/2$. Am I right?
UPDATE: for $m=3$ and $n=4$ we get $-113/2$
UPDATE2: I think $\inf A$ doesn't exist too. I have to prove that e.g:
$$\forall _m \exists _n:m^2n>n^2+2m^2$$
For $n=3$ we have $\frac{m^4+2n^2}{2m^2-m^2n+n^2}\to-\infty$ as $m\to+\infty$, hence $\inf A=-\infty$.