First solution. Rewrite the condition as $x^{2}-x=x(x-1)>y$, $x^{4}-x^{3}=x^{3}(x-1)>y$. We need to prove that $x^{3}-x^{2}=x^{2}(x-1)>y$. Note that $x>1$ - otherwise $x(x-1) \leq 0x(x-1)>y$. Second solution. Since $x^{2}>x+y$ and $x>1, x^{3}>x^{2}+x y>x^{2}+y$. Third solution. Multiply the inequalities $x(x-1)>y$ and $x^{3}(x-1)>y$ (this is possible since $x-1>0$) and take the square root of both sides of the resulting inequality. We obtain the desired inequality $x^{2}(x-1)>y$. Remark. As we can see, the condition $x^{4}>x^{3}+y$ is redundant, but there are solutions that use it.