<s>[INST] To solve $ \dfrac1m+\dfrac1n-\dfrac1{mn^2}=\dfrac34$ How do we find all positive integers $(m,n)$ such that $ \dfrac1m+\dfrac1n-\dfrac1{mn^2}=\dfrac34$ ? 
 [/INST] Since you need $\frac 1m+\frac 1n\gt \frac 34$ you need either $\frac 1m\ge \frac 38\gt \frac 13$ or alternatively $\frac 1n \gt \frac 13$
The number of cases is small and they can be checked quickly.

To be more concrete about the cases involved, we must have either $m\leq 2$ or $n\leq 2$. 
$m=1$ gives $\frac 1n-\frac 1{n^2}=-\frac 14$, or $4(n-1)=-n^2$which is impossible for positive integer $n$. Note that $n^2+4n-4=0$ has roots $n=-2\pm 2\sqrt 2$.
$m=2$ gives $\frac 1n-\frac 1{2n^2}=\frac 14$, which becomes $4n-2=n^2$ and this is impossible in integers since an even square is divisible by $4$. Also $n^2-4n+2=0$ has roots $n=2\pm \sqrt 2$
$n=1$ gives $\frac 1m+1-\frac 1m=\frac 34$ which is false.
$n=2$ gives $\frac 1m-\frac 1{4m}=\frac 14$ or $m=3$
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