Given $x, y$ that $xy-\frac{x}{y^2}-\frac{y}{x^2}=3$, work out $xy-x-y$.

Write $p=xy$, $s=x+y$.

Then the hypothesis rewrites as $p^3-s(s^2-3p)=3p^2$, in other words $(p-s)(p^2+ps+s^2)=p^3-s^3=3p^2-3sp=3p(p-s)$.

So either $p-s=0$, or $3p=p^2+ps+s^2$.

If $p-s \neq 0$, then $3p = 3s^2/4 + (s/2+p)^2$. Since by IAG, $p \leq s^2/4$, then $p=-s/2$ and $x=y$, so $x^2=-x$ and since $x \neq 0$, $x=-1$ and $p-s=3$.

As a conclusion: either $xy-x-y=0$, or $x=y=-1$ and $xy-x-y=3$.