$x, y$ and $z$ - ˙3.0

Let $xy+xz+yz=c$ Then $x+y+z=0$ $xy+xz+yz=c$ $xyz=2$ Since the Vieta's formula the x, y, z are roots of the $t^3+ct-2$ polinom. (Since they can't be all equal.) So $x^3=-cx+2\Rightarrow x^3y=-cxy+2y$ and similarly $y^3z=-cyz+2z$ $z^3x=-czx+2x$ Adding the equations: $x^3y+y^3z+z^3x=-c(xy+xz+yz)+2(x+y+z)=-c^2$ On the other hand $c=xy+xz+yz=x(y+z)+\dfrac{2}{x}=-x^2+\dfrac{2}{x}$ $c<0$ has to be true, because if c\geq0, then $t^3+ct-2$ function is strictly monotonous, so there can not be three real zero positions between which there are different. So we can suppose $x<0$ , and it is enough to determine the minimum of $a^2+\dfrac{2}{a}$ where $a>0$ .

Since the relationship between arithmetic mean and geometric mean: $a^2+\dfrac{2}{a}=a^2+\dfrac{1}{a}+\dfrac{1}{a}\geq3*\sqrt[3]{a^2*\dfrac{1}{a}*\dfrac{1}{a}}=3$ , so its minimum value is $3$ , where $a=1$ . From that $x^3y+y^3z+z^3x\leq -9$ and the equality will only be true, if from the $x, y, z$ two are $-1$ , the third is $2$ .