4. 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: Obviously, the solution of the form $(0, y)$ is unique, which is $(0,0)$. Now we find the solution $(x, y)$ where $x \neq 0$. (2) Multiply by $x$ and subtract (1) to get $\left(2+x^{3}\right) y = x^{2}$. Thus, $2+x^{3} \neq 0$, meaning $y = x^{2} / \left(2+x^{3}\right)$. Substituting the obtained expression for $y$ into (2) and canceling $x$, after equivalent transformations, we get the equation $3 x^{6} + 11 x^{3} + 8 = 0$. Let $x^{3} = z$, solve the equation $3 z^{2} + 11 z + 8 = 0$, yielding $z_{1} = -1, z_{2} = -8 / 3$, i.e., $x_{1} = -1, x_{2} = -2 / \sqrt[3]{3}$. Correspondingly, $y_{1} = 1$, $\boldsymbol{u}_{2} = -2 \sqrt[3]{3}$.

Therefore, the given system of equations has three solutions: $(0,0)$, $(-1,1)$, $(-2 / \sqrt[3]{3}, -2 \sqrt[3]{3})$.