4-106 Solve the system of equations
$$
\left\{\begin{array}{l}
x y^{2}-2 y+3 x^{2}=0 \\
y^{2}+x^{2} y+2 x=0
\end{array}\right.
$$
[Solution] When $x=0$, the system of equations has a unique solution:
$$
\left\{\begin{array}{l}
x=0 \\
y=0
\end{array}\right. \text {. }
$$

When $x \neq 0$, by (1) - (2) $\times x$ we get
$$
-2 y-x^{3} y+x^{2}=0 \text {, }
$$

which simplifies to $\left(2+x^{3}\right) y=x^{2}$,
from which we know $2+x^{3} \neq 0$, so $y=\frac{x^{2}}{2+x^{3}}$.
Substituting into (2) and simplifying, we get $\quad 3 x^{6}+11 x^{3}+8=0$.
Let $x^{3}=z$, solving the equation $3 z^{2}+11 z+8=0$, we get
$$
z_{1}=-1, z_{2}=-\frac{8}{3} \text {. }
$$

Thus,
$$
x_{1}=-1, x_{2}=-\frac{2}{\sqrt[3]{3}} \text {. }
$$

Therefore, $y_{1}=1, y_{2}=-2 \sqrt[3]{3}$. Hence, the system of equations has three solutions:
$$
\left\{\begin{array} { l } 
{ x _ { 1 } = - 1 , } \\
{ y _ { 1 } = 1 ; }
\end{array} \left\{\begin{array} { l } 
{ x _ { 2 } = - \frac { 2 } { \sqrt [ 3 ] { 3 } } , } \\
{ y _ { 2 } = - 2 \sqrt [ 3 ] { 3 } ; }
\end{array} \quad \left\{\begin{array}{l}
x_{3}=0, \\
y_{3}=0 .
\end{array}\right.\right.\right.
$$