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

Translate the text above into English, keeping the original text's line breaks and format, and output the translation result directly. 

The translation is as follows:

Find the range of the function $y=\frac{1}{(x-1)(2 x-1)}$.
"Solve for $x$ first, from the original equation we get
$$
\begin{array}{l}
21 x_{2}^{2}-3 y x+y-1=0, \\
\therefore \quad x=3 y \pm \sqrt{y(y+8)} .
\end{array}
$$

Since $x$ is a real number, we have
$$
\left\{\begin{array}{l}
y \neq 0, \\
y(y+8) \geqslant 0 .
\end{array}\right.
$$

Solving this, we get $y \leqslant-8$ or $y>0$.
Therefore, the range of the function is $\{y \mid y \leqslant-8$ or $y>0\}$." The author feels that the transition from (1) to (2) is not appropriate, and it should be changed to:
From the original equation, we get $2 y x^{2}-3 y x+y-1=0$, (1) when $y \neq 0$,
$$
x=\frac{3 y \pm \sqrt{y(y+8)}}{4 y} .
$$
Since $x$ is a real number,
$$
\therefore\left\{\begin{array}{l}
y(y+8) \geqslant 0, \\
y \neq 0 .
\end{array}\right.
$$

We get $y \leq-8$ or $y>0$.
When $y=0$, (1) becomes $-1=0$, which indicates that $y=0$ is not in the range of the function.
In summary, the range of the function is $\{y \mid y \leqslant-8$ or $y>0\}$