6- 166 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:
(i) $f(x f(y)) \cdot f(y)=f(x+y)$;
(ii) $f(2)=0$;
(iii) $f(x) \neq 0$, when $0 \leqslant x<2$.
[Solution] Let $f$ satisfy the given conditions.
When $x \geqslant 2$, we have
$$\begin{aligned}
f(x) & =f((x-2)+2) \\
& =f((x-2) f(2)) \cdot f(2) \\
& =0
\end{aligned}$$

Thus, $f(x)=0$;
If $0 \leqslant x<2$, we set $y \geqslant 2-x$. Then
$$\begin{array}{l}
x+y \geqslant 2 \\
f(x+y)=0 \\
f(y f(x)) \cdot f(x)=f(y+x)=f(x+y)=0
\end{array}$$

By the given condition, $f(x) \neq 0$ when $0 \leqslant x<2$.
Therefore,
$$\begin{array}{l}
f(y f(x))=0 \\
y f(x) \geqslant 2 \\
y \geqslant \frac{2}{f(x)}
\end{array}$$

This means that from $y \geqslant 2-x$, we get $y \geqslant \frac{2}{f(x)}$. But by the reversibility of the above reasoning, from $y \geqslant \frac{2}{f(x)}$, we can also get $y \geqslant 2-x$. Therefore, we have
$$\begin{array}{l}
2-x=\frac{2}{f(x)}, \\
f(x)=\frac{2}{2-x}, \quad 0 \leqslant x<2 \text { when. }
\end{array}$$

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