Without the loss of generality we assume that $x \ge y \ge z$. Let $x-y = a$ and $y-z = b$. We have to find the maximum value of $ab(a+b)$. Since $x+y+z=0$ and $x^2+y^2+z^2=6$, we get $a^2+b^2+(a+b)^2 = (x-y)^2+(y-z)^2+(x-z)^2 = 3(x^2+y^2+z^2)-(x+y+z)^2 = 18$. Thus, $a^2+ab+b^2 = 9$. By AM-GM inequality,
$$ab \le \frac{(a+b)^2}{4} \le \frac{a^2+ab+b^2}{3} = 3$$
and hence
$$ab(a+b) \le 3 \cdot 2\sqrt{3} = 6\sqrt{3}.$$
Equality holds for $x = \sqrt{3}$, $y = 0$, $z = -\sqrt{3}$.