9. (14th All-Russian Mathematical Olympiad Problem) Solve the system of equations
$$
\left\{\begin{array}{l}
x y^{2}-2 y+3 x^{2}=0, \\
y^{2}+x^{2} y+2 x=0 .
\end{array}\right.
$$
9. When $x=0$, the equation system has a unique solution $(0,0)$.

When $x \neq 0$, subtract the second equation of the original system multiplied by $x$ from the first equation, we get $-2 y - x^{3} y + x^{2} = 0$, which simplifies to $\left(2 + x^{3}\right) y = x^{2}$. From this, we know that $2 + x^{3} \neq 0$, so $y = \frac{x^{2}}{2 + x^{3}}$. Substituting this into the second equation of the original system, we get $3 x^{6} + 11 x^{3} + 8 = 0$. Let $y^{3} = z$, solving the equation $3 z^{2} + 11 z + 8 = 0$ yields $z_{1} = -1, z_{2} = -\frac{8}{3}$. Thus, $x_{1} = -1, x_{2} = -\frac{2}{\sqrt[3]{3}}$, and therefore $y_{1} = 1, y_{2} = -2 \sqrt[3]{x}$. Hence, the solutions $(x, y)$ of the original equation system are $(-1,1),\left(-\frac{2}{\sqrt[3]{3}},-2 \sqrt[3]{3}\right),(0,0)$.