$\therefore$ Let $x, y, z \geqslant 0$, and $x+y+z=1$, find the maximum and minimum value of $\iota y+y z+z x-3 x y z$.
13. Let $x \geqslant y \geqslant z$, then $x \geqslant \frac{1}{3}, z \leqslant \frac{1}{3}$. Therefore, $x y+y z+z x-3 x y z=x y(1-3 z)+ y z+z x \geqslant 0$, and the equality holds when $x=1, y=z=0$. The minimum value is 0.
Let $z=\frac{1}{3}-k$, then $0 \leqslant k \leqslant \frac{1}{3}$, thus
$$
\begin{aligned}
& x y+y z+z x-3 x y z=z(x+y)+x y(1-3 z) \\
= & \left(\frac{1}{3}-k\right)\left(\frac{2}{3}+k\right)+x y \cdot(3 k) \\
\leqslant & \left(\frac{1}{3}-k\right)\left(\frac{2}{3}+k\right)+\frac{(x+y)^{2}}{4} \cdot 3 k \\
= & \left(\frac{1}{3}-k\right)\left(\frac{2}{3}+k\right)+\frac{1}{4}\left(\frac{2}{3}+k\right)^{2} \cdot 3 k \\
= & \frac{2}{9}+\frac{3}{4} k^{4} \leqslant \frac{2}{9}+\frac{3}{4} \cdot\left(\frac{1}{3}\right)^{3} \\
= & \frac{1}{4}
\end{aligned}
$$

When $x=y=-\frac{1}{2}, z=0$, the equality holds. Hence, the maximum value is $\frac{1}{4}$.