F9 (27-5, UK) Let $f$ be a function defined on the set of non-negative real numbers and taking values in the same set. Find all functions $f$ that satisfy the following conditions:
(1) $f(x f(y)) f(y)=f(x+y)$;
(2) $f(2)=0$;
(3) $f(x) \neq 0$, when $0 \leqslant x<2$.
Let $y=2$ be substituted into (1), we get
$$
f(x+2)=0 .
$$

Since $x \geqslant 0$, it follows from (1) that when $x \geqslant 2$, $f(x)=0$. Furthermore, from (3), when $f(x)=0$, $x \geqslant 2$.
Also, when $f(x f(y))=0$, by (1) we have $f(x+y)=0$, which means that when $x+y \geqslant 2$, either $y \geqslant 2$ or $x f(y) \geqslant 2$.
Therefore, when $0 \leqslant y<2$, $x f(y) \geqslant 2$ is equivalent to $x+y \geqslant 2$. That is,
$$
x \geqslant \frac{2}{f(y)} \Leftrightarrow x \geqslant 2-y,
$$

Thus,
$$
\begin{array}{l}
\frac{2}{f(y)}=2-y, \\
f(y)=\frac{2}{2-y} .
\end{array}
$$

In summary, the function sought is
$$
f(x)=\left\{\begin{array}{ll}
\frac{2}{2-x}, & \text { when } 0 \leqslant x<2 \\
0, & \text { when } x \geqslant 2 \text { . }
\end{array}\right.
$$