Example 2 Let $0<x<\frac{9}{2}$. Find the range of the function
$$
y=\left[1+\frac{1}{\lg \left(\sqrt{x^{2}+10}+x\right)}\right]\left[1+\frac{1}{\lg \left(\sqrt{x^{2}+10}-x\right)}\right]
$$
Given $0<x<\frac{9}{2}$, we know
$$
0<\lg \left(\sqrt{x^{2}+10}-x\right)<\lg \left(\sqrt{x^{2}+10}+x\right)<1 .
$$

Notice that,
$$
\lg \left(\sqrt{x^{2}+10}+x\right)+\lg \left(\sqrt{x^{2}+10}-x\right)=1 \text {. }
$$

We can set $\lg \left(\sqrt{x^{2}+10}+x\right)=\sin ^{2} \theta$,
$$
\lg \left(\sqrt{x^{2}+10}-x\right)=\cos ^{2} \theta \text {, }
$$

where $0<\theta<\frac{\pi}{4}$.
$$
\begin{array}{l}
\text { Then } y=\left(1+\frac{1}{\sin ^{2} \theta}\right)\left(1+\frac{1}{\cos ^{2} \theta}\right) \\
=\left(2+\frac{1}{\tan ^{2} \theta}\right)\left(2+\tan ^{2} \theta\right) \\
=5+2 \tan ^{2} \theta+\frac{2}{\tan ^{2} \theta} .
\end{array}
$$

Let $t=\tan ^{2} \theta(0<t<1)$. It is easy to get $y^{\prime}=2-\frac{2}{t^{2}}<0$.
Therefore, $y=5+2 t+\frac{2}{t}$ is monotonically decreasing on $0<t<1$.
When $t \rightarrow 0$, $y \rightarrow+\infty$;
When $t \rightarrow 1$, $y \rightarrow 9$.
Thus, the range of the function $y$ is $(9,+\infty)$.