13. Let real numbers $x, y, z$ satisfy $x+y+z=0$, and $(x-y)^{2}+(y-z)^{2}+(z-x)^{2} \leqslant 2$. Find the maximum and minimum values of $x$.
13. From the problem, we know
$$
\begin{aligned}
2 & \geqslant(x-y)^{2}+(y-z)^{2}+(z-x)^{2} \\
= & 2 x^{2}+2 y^{2}+2 z^{2}-2 x y-2 y z-2 z x \\
= & 2 x^{2}+2(y+z)^{2}-4 y z-2(y+z)-2 y z \\
& 6 x^{2}-6 y z \\
& =6 x^{2}-6 \cdot \frac{(y+z)^{2}-(y-z)^{2}}{4} \\
& =\frac{9}{2} x^{2}+\frac{3}{2}(y-z)^{2} \geqslant \frac{9}{2} x^{2} .
\end{aligned}
$$

Thus, $-\frac{2}{3} \leqslant x \leqslant \frac{2}{3}$.
When $x=-\frac{2}{3}, y=z=\frac{1}{3}$, the conditions of the problem are satisfied; when $x=\frac{2}{3}, y=z=-\frac{1}{3}$, the conditions of the problem are also satisfied. Therefore, the maximum value of $x$ is $\frac{2}{3}$, and the minimum value is $-\frac{2}{3}$.