Example 3 Let $0<a \neq 1$ and $f(x)=\log _{a}\left(6 a x^{2}-2 x+3\right)$ be monotonically increasing on $\left[\frac{3}{2}, 2\right]$, find the range of values for $a$.

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Analyzing, we should divide into $a>1$ and $0<a\leq1$. From $a x^{2}-\frac{1}{3} x+\frac{1}{2}>0$ we know: $a>-\frac{1}{2 x^{2}}+\frac{1}{3 x}=-\frac{1}{2}\left(\frac{1}{x}-\frac{1}{3}\right)^{2}+\frac{1}{18}$ always holds. Since $\frac{3}{2} \leqslant x \leqslant 2$, i.e., $\frac{1}{2} \leqslant \frac{1}{x} \leqslant \frac{2}{3}$, we get $: a>\frac{1}{24}$.
(1) When $\frac{1}{24}<a\leq1$, $f(x)$ is monotonically increasing on $\left[\frac{3}{2}, 2\right]$, which is equivalent to $g(x)=6 a x^{2}-2 x+3$ being monotonically increasing on $\left[\frac{3}{2}, 2\right]$, then $x_{0}=\frac{1}{6 a} \leqslant \frac{3}{2} \Rightarrow a \geqslant \frac{1}{9}$, so $a \in\left(\frac{1}{24}, \frac{1}{12}\right]$.
(2) When $a>1$, $f(x)$ is monotonically increasing on $\left[\frac{3}{2}, 2\right]$, which is equivalent to $g(x)=6 a x^{2}-2 x+3$ being monotonically increasing on $\left[\frac{3}{2}, 2\right]$, then $x_{0}=\frac{1}{6 a} \leqslant \frac{3}{2} \Rightarrow a \geqslant \frac{1}{9}$, hence $a>1$.
In summary, $a \in\left(\frac{1}{24}, \frac{1}{12}\right] \cup(1,+\infty)$.
/Note: The condition that the logarithm is greater than 0, i.e., $a>\frac{1}{24}$, is easily overlooked, and should be noted.