Example 4 Let $f: \mathbf{R}^{*} \rightarrow \mathbf{R}^{*}$, and satisfy
(1) $f(x f(y)) f(y)=f(x+y)$;
(2) $f(2)=0$;
(3) $f(x) \neq 0$ for all $0 \leqslant x<2$.

Try to find the function $f(x)$ that satisfies the conditions.
(27th IMO)
Analysis: From conditions (1), (2), and $x=x-2+2$, repeatedly applying condition (1) can determine the properties that $f(x)$ satisfies, and then find $f(x)$.
When $x>2$,
$$
\begin{array}{l}
f(x)=f((x-2)+2) \\
=f((x-2) f(2)) f(2)=0 .
\end{array}
$$

When $0 \leqslant x < 2$, if $f(x) > 0$, then
$$
\begin{array}{l}
f(x) f((2-x) f(x)) \\
=f((2-x)+x)=f(2)=0 .
\end{array}
$$

By (3), we have
$$
f((2-x) f(x))=0.
$$

Therefore, $(2-x) f(x) \geqslant 2$, which means $f(x) \geqslant \frac{2}{2-x}$.
Also, when $0 \leqslant x < 2$, we have
$$
\begin{aligned}
0 & =f(2)=f\left(\frac{2}{f(x)} \cdot f(x)\right) f(x) \\
& =f\left(\frac{2}{f(x)}+x\right),
\end{aligned}
$$

Thus, $\frac{2}{f(x)}+x \geqslant 2$.
Hence, $f(x) \leqslant \frac{2}{2-x}$.
Therefore, $f(x)=\left\{\begin{array}{ll}\frac{2}{2-x}, & 0 \leqslant x<2, \\ 0 & x \geqslant 2 .\end{array}\right.$
It is easy to verify that $f(x)$ satisfies conditions (1), (2), and (3).