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. $\left\{x \left\lvert\,-\frac{2}{3} \leqslant x<0\right.\right\}$.

Let $y=\sqrt[3]{3 x+1}$, then $3 x+1=y^{3}, 3 x=y^{3}-1$. The original inequality transforms to $\frac{(3 x)^{2}}{(1-\sqrt[3]{3 x+1})^{2}} \leqslant 3 x + y \leqslant 1$, which simplifies to $-1 \leqslant \sqrt[3]{3 x+1} \leqslant 1$, yielding $-\frac{2}{3} \leqslant x \leqslant 0$. Also, $1-\sqrt[3]{3 x+1} \neq 0$, so $x \neq 0$. Therefore, $-\frac{2}{3} \leqslant x < 0$