Given $x, y$ that $xy-\frac{x}{y^2}-\frac{y}{x^2}=3$, work out $xy-x-y$. Today I had a competition in Xiamen, China. I know how to do the questions except this strange equation.

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

Such a strange question, right? I have found the integral solution are $0$ when $x=y=2$ and $3$ when $x=y=-1$, but I think there are infinitely many solutions but I can’t really prove it at that time. Can you guys help me?

Making $y = \lambda x$ we have
$$
xy -\frac{x}{y^2}-\frac{y}{x^2}=3\Rightarrow \lambda x^3-3x=\lambda+\frac{1}{\lambda^2}
$$
and solving for $x$
$$
x = \left\{
\begin{array}{c}
 \frac{\lambda +1}{\lambda } \\
 -\frac{\lambda ^2+\lambda +\sqrt{3} \sqrt{-(\lambda -1)^2 \lambda ^2}}{2 \lambda ^2} \\
 -\frac{\lambda ^2+\lambda -\sqrt{3} \sqrt{-(\lambda -1)^2 \lambda ^2}}{2 \lambda ^2} \\
\end{array}
\right.
$$
so the solutions for $x$ are all real if $\lambda = 1$ and in those circumstances
$$
x = y = \cases{2\\
-1}
$$
giving respectively
$$
x y - x - y = 0\\
x y - x - y = 3
$$
NOTE
$$
\lambda^3 x^3-3\lambda^2x-\lambda^3-1 = (\lambda x-\lambda-1)(\lambda^2x^2+(\lambda^2+\lambda) x+\lambda^2-\lambda+1)
$$
