Example 10. (IMO-27-5)
$f$ is a function defined on the non-negative real numbers and taking non-negative real values. Find all functions $f$ that satisfy the following conditions:
( i ) $f(x f(y)) f(y)=f(x+y)$;
( ii ) $f(2)=0$;
(iii) $f(x) \neq 0$, when $0 \leqslant x<2$.


If $w>2$, then in (i) let $y=2, x=w-2$, we get
$$
f(w)=f((w-2) f(2)) f(2)=0 .
$$

Thus,
$$
f(x)=0 \Leftrightarrow x \geqslant 2 .
$$

In (i), let $x>0, 0 \leqslant y < 2$, then when $x>0$, we have
$$
x f(y) \geqslant 2 \Leftrightarrow x+y \geqslant 2 .
$$

This is equivalent to
$$
\begin{array}{l}
f(y) \geqslant \frac{2}{x} \Leftrightarrow x \geqslant 2-y, \\
f(y)<\frac{2}{x} \Leftrightarrow x<2-y .
\end{array}
$$

Fix $y$, and let $x$ approach $2-y$ from the left and right, from the above two equations we get:
$$
f(y)=\frac{2}{2-y} .
$$

Thus, the function $f$ we seek is
$$
f(y)=\left\{\begin{array}{cl}
\frac{2}{2-y}, & \text { when } 0 \leqslant y<2 ； \\
0 . & \text { when } y \geqslant 2 .
\end{array}\right.
$$

It clearly satisfies conditions (ii) and (iii). Now we verify that it also satisfies condition (i).
1. When $x+y<2$ and $x f(y)<2$, at this time $0 \leqslant y<2, \quad f(y)=\frac{2}{2-y}$
(Author's affiliation: University of Science and Technology of China)