$4 \cdot 110$ Solve the system of equations
$$
\left\{\begin{array}{l}
y^{2}=x^{3}-3 x^{2}+2 x, \\
x^{2}=y^{3}-3 y^{2}+2 y .
\end{array}\right.
$$

for all solutions.
[Solution] (1)-(2) gives
$$
y^{2}-x^{2}=\left(x^{3}-3 x^{2}+2 x\right)-\left(y^{3}-3 y^{2}+2 y\right) \text {, }
$$

which simplifies to $y^{3}-2 y^{2}+2 y=x^{3}-2 x^{2}+2 x$.
Let $f(x)=x^{3}-2 x^{2}+2 x$, then equation (3) becomes
$$
f(y)=f(x) \text {. }
$$

Since
$$
\begin{array}{l}
f^{\prime}(x)=3 x^{2}-4 x+2, \\
\Delta=(-4)^{2}-4 \cdot 3 \cdot 2=-80, \text{ thus the function } f(x) \text{ is strictly increasing.}
\end{array}
$$
Therefore, from (4) we get
$$
x=y \text {. }
$$

Substituting (5) into (1) gives
$$
y^{2}=y^{3}-3 y^{2}+2 y \text {, }
$$

which simplifies to $y^{3}-4 y^{2}+2 y=0$,
$$
\begin{array}{l}
y\left(y^{2}-4 y+2\right)=0 \\
y_{1}=0 ; y_{2,3}=2 \pm \sqrt{2}
\end{array}
$$

From (5), the solutions to the original system of equations are
$$
\left\{\begin{array} { l } 
{ x _ { 1 } = 0 , } \\
{ y _ { 1 } = 0 ; }
\end{array} \quad \left\{\begin{array} { l } 
{ x _ { 2 } = 2 + \sqrt { 2 } , } \\
{ y _ { 2 } = 2 + \sqrt { 2 } ; }
\end{array} \quad \left\{\begin{array}{l}
x_{3}=2-\sqrt{2} \\
y_{3}=2-\sqrt{2}
\end{array}\right.\right.\right.
$$