Solve in positive integers $\frac{x^{2}}{y}+\frac{y^{2}}{x}=9$

By AM-GM, $\frac{x^2}{y}+\frac{y^2}{x}\geq 2\sqrt{xy}$, hence $xy$ is at most $20$.

Moreover, the original equation is equivalent to: $$ (x+y)(x^2-xy+y^2) = 9xy, $$ but $x^2-xy+y^2\geq xy$, hence $x+y$ is at most $9$. So we have to check very few cases to be sure that the only solutions in $\mathbb{N}^2$ are the trivial ones.