$A=\left\{\frac{m^2-n}{m^2+n^2}:m,n\in \Bbb N,m>2n>1\right\} $
$\sup A=1$
$1$ is upper bound: $\frac{m^2-n}{m^2+n^2}\le1\\m^2-n\le m^2+n^2 \\ -n \le n^2 \mbox{ true}$
And for $n=1$,$ \lim \limits_{m \to \infty\ } \frac{m^2-1}{m^2+1}=1$
I suppose infimum A=3/5, because
$\frac{m^2-n}{m^2+n^2}\ge\frac{(2n)^2-n}{(2n)^2+n^2}=\frac{4n^2-n}{5n^2}$, and for $n=1$ it equals $3/5$, but I don't sure it. Could someone help me with $\inf A$? Thank you in advance.
One small correction I would make would be as follows. For all fixed $n\in\mathbb{N}-\left\{1 , 0\right\}$ we have $$ \frac{m^2-n}{m^2+n^2}= \frac{1-\left(\frac{n}{m^2}\right)}{1+\left(\frac{n^2}{m^2}\right)} \leq 1 $$ and $$ \lim_{m\to\infty} \frac{m^2-n}{m^2+n^2} = \lim_{m\to\infty} \frac{1-\frac{n}{m^2}}{1+\frac{n^2}{m^2}} = \frac{1-\lim_{m\to\infty}\left(\frac{n}{m^2}\right)}{1+\lim_{m\to\infty}\left(\frac{n^2}{m^2}\right)} =\frac{1-0}{1+0}=1 $$ His reasoning for the infimum needs more argumentation though you has posted the right answer. You need to argue that the function $\mathbb{N}-\left\{0,1\right\}\ni n\mapsto \frac{4n^2-n}{5n^2}$ is increasing.