<s>[INST] 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?
 [/INST] It is, of course, possible to solve the cubic $$xy-\frac x{y^2}-\frac y{x^2}=3\implies y^3\color{red}{-\frac{3x^2}{x^3-1}}y^2\color{blue}{+0}y\color{green}{-\frac{x^3}{x^3-1}}=0$$ using Cardano's method, which however may not be suitable as a contest problem.
We have $$Q=\frac{3\cdot\color{blue}{0}-\left(\color{red}{-\dfrac{3x^2}{x^3-1}}\right)^2}9=-\frac{x^4}{(x^3-1)^2}\tag1$$ and $$\small R=\frac{9\cdot\left(\color{red}{-\dfrac{3x^2}{x^3-1}}\right)\cdot\color{blue}0-27\cdot\left(\color{green}{-\dfrac{x^3}{x^3-1}}\right)-2\left(\color{red}{-\dfrac{3x^2}{x^3-1}}\right)^3}{54}=\frac{x^3}{2(x^3-1)}\left(1+\frac{2x^3}{(x^3-1)^2}\right)\tag2.$$ Now $$\small\sqrt{Q^3+R^2}=\sqrt{-\frac{x^{12}}{(x^3-1)^6}+\frac{x^6}{4(x^3-1)^2}\left(1+\frac{4x^3}{(x^3-1)^2}+\frac{4x^6}{(x^3-1)^4}\right)}=\frac{x^3(x^3+1)}{2(x^3-1)^2}\tag3$$ and so for $x\ne-1$, \begin{cases}R-\sqrt{Q^3+R^2}=\frac{x^3}{2(x^3-1)^2}\left(x^3-1+\frac{2x^3}{x^3-1}-(x^3+1)\right)=\frac{x^3}{(x^3-1)^3}\\R+\sqrt{Q^3+R^2}=\frac{x^3}{2(x^3-1)^2}\left(x^3-1+\frac{2x^3}{x^3-1}+(x^3+1)\right)=\frac{x^9}{(x^3-1)^3}\tag4.\end{cases} Therefore, $$S=\sqrt[3]{R+\sqrt{Q^3+R^2}}=\frac{x^3}{x^3-1},\quad T=\sqrt[3]{R-\sqrt{Q^3+R^2}}=\frac{x}{x^3-1}\tag5$$ and since $S-T\ne0$, the only real solution is $$y=S+T-\frac13\cdot\left(\color{red}{-\dfrac{3x^2}{x^3-1}}\right)=\frac{x^3+x^2+x}{x^3-1}=\frac x{x-1}=1+\frac1{x-1}\tag6.$$ Finally, this yields $$xy-x-y=x+\frac x{x-1}-x-1-\frac1{x-1}=0.\tag7$$
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