jmc

We can factor the expression as $$xy(x-y).$$Since we are trying to find the maximum, we can assume that $x\ge y.$ Then by AM-GM, $$y(x-y)\le \frac{x^2}{4},$$so $$xy(x-y)\le x\cdot \frac{x^2}{4}=\frac{x^3}{4}\le \frac{1}{4}.$$Equality occurs when $x=1$ and $y=\frac{1}{2},$ giving us a maximum value of $\boxed{\frac{1}{4}}.$