We are given the equation $x^2 y = x - y$, and we want to prove that for every real number $x$, there is a unique real number $y$ that satisfies this equation. We'll analyze the different cases for $x$ and solve for $y$ accordingly.

Case 1: $x = 0$

When $x = 0$, the equation becomes: $0^2 y = 0 - y$ which simplifies to: $0 = -y$ Thus, $y = 0$.

So, for $x = 0$, the unique solution is $y = 0$.

Case 2: $x \neq 0$

When $x \neq 0$, the equation is: $x^2 y = x - y$ We want to solve this equation