Example 8 Find the smallest positive period of the function $f(x)=\sin x \sin \frac{1}{2} x \sin \frac{1}{3} x$.
Solution It is obvious that $12 \pi$ is a period of $f(x)$. To find the smallest positive period $T(0<T<12 \pi)$ of $f(x)$, from $f(x+T)=f(x)$, we get
$$
\sin (x+T) \sin \frac{1}{2}(x+T) \sin \frac{1}{3}(x+T)=\sin x \sin \frac{1}{2} x \sin \frac{1}{3} x .
$$

In the interval $(0, \pi)$, $\sin x \neq 0, \sin \frac{1}{2} x \neq 0, \sin \frac{1}{3} x \neq 0$,
In (1), let $x \rightarrow 0$, we get $\sin T \sin \frac{1}{2} T \sin \frac{1}{3} T=0$,