jmc

Let $k=xy+xz+yz.$ Then by Vieta's formulas, $x,$ $y,$ and $z$ are the roots of $$t^3+kt-2=0.$$Then $x^3+kx-2=0,$ so $x^3=2-kx,$ and $x^{3}y=2y-kxy.$ Similarly, $y^{3}z=2z-kyz$ and $z^{3}x=2x-kxz,$ so $$x^{3}y+y^{3}z+z^{3}x=2(x+y+z)-k(xy+xz+yz)=-k^2.$$Since $xyz=2,$ none of $x,$ $y,$ $z$ can be equal to 0. And since $x+y+z=0,$ at least one of $x,$ $y,$ $z$ must be negative. Without loss of generality, assume that $x<0.$ From the equation $x^3+kx-2=0,$ $x^2+k-\frac{2}{x}=0,$ so $$k=\frac{2}{x}-x^2.$$Let $u=-x,$ so $u>0,$ and $$k=-\frac{2}{u}-u^2=-\left(u^2+\frac{2}{u}\right).$$By AM-GM, $$u^2+\frac{2}{u}=u^2+\frac{1}{u}+\frac{1}{u}\ge 3\sqrt[3]{u^2\cdot \frac{1}{u}\cdot \frac{1}{u}}=3,$$so $k\le -3$. Therefore, $$x^{3}y+y^{3}z+z^{3}x=-k^2\le -9.$$Equality occurs when $x=y=-1$ and $z=2,$ so the maximum value is $\boxed{-9}.$