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 $\sin T \sin \frac{1}{2} T \sin \frac{1}{3} T=0$,
Equivalent to $\sin T=0$, so $T=k_{1} \pi\left(0<k_{1} \leqslant 12\right)$. In the interval $(0, \pi)$, since $\sin (x+T)=(-1)^{k} \sin x \neq 0$,
from (1) we get
$$
\sin \frac{1}{2}(x+T) \sin \frac{1}{3}(x+T)=\sin \frac{1}{2} x \sin \frac{1}{3} x \text {, }
$$

Let $x \rightarrow 0$, we get $\sin \frac{1}{2} T \sin \frac{1}{3} T=0, \sin \frac{1}{2} T$ and $\sin \frac{1}{3} T$ at least one is zero. Without loss of generality, let $\sin \frac{1}{2} T=0, T=4 k_{2} \pi, k_{2} \in \mathbf{Z}$. Using a similar method as before, we get for $x \in(0, \pi)$
$$
\sin \frac{1}{3}(x+T)=\sin \frac{1}{3} x,
$$

Let $x \rightarrow 0$, we get
$$
\sin \frac{1}{3} T=0, T=6 k_{3} \pi, k_{3} \in \mathbf{Z},
$$

Since $T=k_{1} \pi=4 k_{2} \pi=6 k_{3} \pi$; hence $T$ must be the least common multiple of $\pi, 4 \pi, 6 \pi$, which is $12 \pi$, or its multiples $24 \pi, 36 \pi, \cdots$
Upon verification, $12 \pi$ is the smallest positive period of the given function.