jmc

Without loss of generality, we can assume that $z\le x$ and $z\le y.$ Then $$(x^2-xy+y^2)(x^2-xz+z^2)(y^2-yz+z^2)\le (x^2-xy+y^2)x^2{y}^2.$$By AM-GM, $$\begin{array}{rl}{x}^2{y}^2(x^2-xy+y^2)& =\frac{4}{9}\left(\frac{3}{2}xy\right)\left(\frac{3}{2}xy\right)(x^2-xy+y^2)\\ & \le \frac{4}{9}{\left(\frac{\frac{3}{2}xy+\frac{3}{2}xy+(x^2-xy+y^2)}{3}\right)}^3\\ & =\frac{4}{9}{\left(\frac{x^2+2xy+y^2}{3}\right)}^3\\ & =\frac{4}{9}\cdot \frac{(x+y{)}^6}{27}\\ & \le \frac{4}{243}(x+y+z{)}^6\\ & =\frac{256}{243}.\end{array}$$Equality occurs when $x=\frac{4}{3},$ $y=\frac{2}{3},$ and $z=0,$ so the maximum value is $\boxed{\frac{256}{243}}.$