4・138 Solve the system of equations $\left\{\begin{array}{l}x^{2}+2 y z=x, \\ y^{2}+2 z x=z, \\ z^{2}+2 x y=y .\end{array}\right.$
[Solution] Adding the three equations yields $(x+y+z)^{2}=x+y+z$
Therefore, $\quad x+y+z=0$ or $x+y+z=1$
If $x+y+z=0$, then $x=-(y+z)$,

Substituting (4) into (2) gives $y^{2}-2 z(y+z)=z$,
Substituting (4) into (3) gives $z^{2}-2 y(y+z)=y$,
(3) - (6) yields $3\left(y^{2}-z^{2}\right)=z-y$.

Thus, $\quad y=z$ or $y+z=-\frac{1}{3}$.
When $y=z$, from (5) we get $y^{2}-4 y^{2}=y$, i.e., $y=0,-\frac{1}{3}$.
Thus, we have $\approx=0,-\frac{1}{3}, x=0, \frac{2}{3}$.
When $y+z=-\frac{1}{3}$, from (4) we get $x=\frac{1}{3}$, substituting into (1) yields $y z=\frac{1}{9}$.

So, $y=\frac{-1 \pm \sqrt{3} i}{6}, z=\frac{-1 \mp \sqrt{3} i}{6}$.
If $x+y+z=1$, then $x=1-y-z$, substituting into (2), (3) and simplifying gives
$$
\begin{array}{l}
y^{2}-2 z^{2}-2 y z+z=0 \\
z^{2}-2 y^{2}-2 y z+y=0
\end{array}
$$
(7) - (8) yields $3\left(y^{2}-z^{2}\right)=y-\approx$.

Thus, $y=z$ or $y+z=\frac{1}{3}$.
When $y=z$, from (7) we get $3 y^{2}=y$, i.e., $y=0, \frac{1}{3}$,

So, $z=0, \frac{1}{3} ; x=1, \frac{1}{3}$.
When $y+z=\frac{1}{3}$, $x=\frac{2}{3}$, from (1) we get $y z=\frac{1}{9}$,
So, $y=\frac{1 \pm \sqrt{3} i}{6}, z=\frac{1 \mp \sqrt{3} i}{6}$.
In summary, the solutions to this problem are
$$
\left\{\begin{array}{l}
x=0, \frac{2}{3}, \frac{1}{3}, 1, \frac{1}{3}, \frac{2}{3} ; \\
y=0,-\frac{1}{3}, \frac{-1 \pm \sqrt{3} i}{6}, 0, \frac{1}{3}, \frac{1 \pm \sqrt{3} i}{6} \\
z=0,-\frac{1}{3}, \frac{-1 \mp \sqrt{3} i}{6}, 0, \frac{1}{3}, \frac{1 \mp \sqrt{3} i}{6}
\end{array}\right.
$$