3. The solution set of the inequality $\frac{3 x^{2}}{(1-\sqrt[3]{3 x+1})^{2}} \leqslant x+2+$ $\sqrt[3]{3 x+1}$ is $\qquad$
3. $-\frac{2}{3} \leqslant x<0$.

Let $y=\sqrt[3]{3 x+1}$. Then $3= \pm i=y^{3}, 3 x=y^{3}$, the original inequality becomes
$$
\begin{array}{l}
\frac{(3 x)^{2}}{1-\sqrt[3]{3}-x+1)^{2}} \leqslant 3 x+6+3 \sqrt[3]{3 x+1}, \\
\frac{\left(y^{3}-1\right)^{2}}{(1-y)^{2}} \leqslant y^{3}+5+3 y, \\
\left(y^{2}+y+1\right)^{2} \leqslant y^{3}+5+3 y .
\end{array}
$$

Simplifying and organizing, we get
$$
\begin{array}{l}
y^{4}+y^{3}+3 y^{2}-y-4 \leqslant 0, \\
(y+1)(y-1)\left(y^{2}+y+4\right) \leqslant 0 .
\end{array}
$$
Since $y^{2}+y+4$ is always positive,
$$
\begin{array}{l}
\therefore-1 \leqslant y \leqslant 1, \\
-1 \leqslant \sqrt[3]{3 x+1} \leqslant 1 .
\end{array}
$$

Solving, we get $-\frac{2}{3} \leqslant x \leqslant 0$.
$$
\begin{array}{l}
\text { Also, } \because 1-\sqrt[3]{3 x+1} \neq 0, \\
\therefore x \neq 0 .
\end{array}
$$

Therefore, the solution set of the original inequality is: $-\frac{2}{3} \leqslant x<0$.