jmc

The condition is equivalent to $z^2+(x+y-1)z=0$. Since $z$ is positive, $z=1-x-y$, so $x+y+z=1$. By the AM-GM inequality, $$xyz\le {\left(\frac{x+y+z}{3}\right)}^3=\boxed{\frac{1}{27}},$$with equality when $x=y=z=\frac{1}{3}$.