(Example 10 Given a function $f(x)$ defined on $\mathbf{R}^{+}$, for any positive numbers $x_{1}$ and $x_{2}$, it holds that $f\left(x_{1} x_{2}\right)=$ $f\left(x_{1}\right)+f\left(x_{2}\right)$, and $f(8)=3$. Find the value of $f(\sqrt{2})$.
From the given information, we have $f(2 \times 2)=f(2)+f(2), f(4 \times 2)=f(4)+f(2)$, thus, $f(8)=3 f(2)$, so $3 f(2)=3$, solving for $f(2)=1$.

Also, $f(\sqrt{2} \times \sqrt{2})=f(\sqrt{2})+f(\sqrt{2})$, hence $f(\sqrt{2})=\frac{1}{2}$.