3. The solution set of the inequality $\frac{4 x^{2}}{(1-\sqrt{1+2 x})^{2}}<2 x+9$ is
3. $-\frac{1}{2} \leqslant x<0$ or $0<x<\frac{45}{8}$.

The original inequality can be simplified to
$$
x^{2}(8 x-45)<0 \Rightarrow x<\frac{45}{8} \text {. }
$$

From $1+2 x \geqslant 0$, and $1-\sqrt{1+2 x} \neq 0$ $\Rightarrow x \geqslant-\frac{1}{2}$, and $x \neq 0$.
Therefore, $-\frac{1}{2} \leqslant x<0$ or $0<x<\frac{45}{8}$.