$x, y$ and $z$

We will show that the maximum value of the expression is $\frac{4}{27}$ .

Suppose $x$ is the largest (or one of the) number from $x, y, z$ . Now $y^2z\leq xyz$ and $z^2\leq\frac{zx}{2}+\frac{z^2}{2}$ , so

$x^2y+y^2z+z^2x\leq x^2y+xyz+z^2x\leq x(x+z)(y+\frac{z}{2})=\frac{1}{2}x(x+z)(2y+z)\leq\frac{1}{2}(\frac{2(x+y+z)}{3})^3=\frac{4}{27}$

(We used the the relationship between arithmetic mean and geometric mean.)

The equality will only be true, if $z=0$ and $x=2y$ , so $z=0$ , $y=\frac{1}{3}$ , $x=\frac{2}{3}$ .