Solving the three equations by adding them yields $(x+y+z)^{2}=x+y+z$,
Therefore, $x+y+z=0$ or $x+y+z=1$.
(1) If $x+y+z=0$, then $x=-(y+z)$.
Substituting (4) into (2), we get $y^{2}-2 z(y+z)=z$,
Substituting (4) into (3), we get $z^{2}-2 y(y+z)=y$,
(5) - (6), we get $3\left(y^{2}-z^{2}\right)=z-y$.

Thus, we find $y=z$ 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 find $z=0,-\frac{1}{3} ; x=0, \frac{2}{3}$.
When $y+z=-\frac{1}{3}$, from (4) we get $x=\frac{1}{3}$, substituting into (1) we get $y z=\frac{1}{9}$, so $y=\frac{1}{6}(-1 \pm \sqrt{3} \mathrm{i}), z=\frac{1}{6}(-1 \mp \sqrt{3} \mathrm{i})$.
(2) If $x+y+z=1$, then $x=1-y-z$, substituting into (2), (3) we get
$y^{2}-2 z^{2}-2 y z+z=0$,
$z^{2}-2 y^{2}-2 y z+y=0$,
From (7) - (8), we get $3\left(y^{2}-z^{2}\right)=y-z$,
Thus, $y=z$ or $y+z=\frac{1}{3}$.
When $y=z$, from (7) we get $3 y^{2}=y$, i.e., $y=0, \frac{1}{3}$.
At this time, $z=0, \frac{1}{3} ; x=1, \frac{1}{3}$.
When $y+z=\frac{1}{3}$, $x=\frac{2}{3}$, from (1) we get $y z=\frac{1}{9}$.
At this time, $y=\frac{1}{6}(1 \pm \sqrt{3} \mathrm{i}), z=\frac{1}{6}(1 \mp \sqrt{3} \mathrm{i})$.
In summary, the solutions $(x, y, z)$ for this problem are $(0,0,0),\left(\frac{2}{3},-\frac{1}{3},-\frac{1}{3}\right),\left(\frac{1}{3}, \frac{-1 \pm \sqrt{3} \mathrm{i}}{6}\right.$, $\left.\frac{-1 \mp \sqrt{3} i}{6}\right),(1,0,0),\left(\frac{1}{3}, \frac{1}{3}, \frac{1}{3}\right),\left(\frac{2}{3}, \frac{1 \pm \sqrt{3} i}{6}, \frac{1 \mp \sqrt{3} i}{6}\right)$.
Example 21 (Canada National Training Team Practice Question) Solve the system of equations in the set of complex numbers:
$$
\left\{\begin{array}{l}
x^{2}+2 y z=x, \\
y^{2}+2 z x=z, \\
z^{2}+2 x y=y .
\end{array}\right.
$$