2. The range of the function $y=\frac{1}{(x-1)(2 x-1)}$ is

保留了源文本的换行和格式，这是翻译结果。
2. $\{y \mid y \leqslant-8$ or $y>0\}$.
$$
y=\frac{1}{2 x^{2}-3 x+1}=\frac{1}{2\left(x-\frac{3}{4}\right)^{2}-\frac{1}{8}} \text {, }
$$

i.e., $16\left(x-\frac{3}{4}\right)^{2} y-y=8$.
$$
\therefore\left(x-\frac{3}{4}\right)^{2}=\frac{8+y}{16 y} \geqslant 0 \text {, }
$$

i.e., $y(y+8) \geqslant 0$, and $y \neq 0$.
Thus, $y>0$ or $y \leqslant-8$,

i.e., $\{y \mid y \leqslant-8$ or $y>0\}$ is the solution.