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
$$
\begin{aligned}
f(x) & =f((x-2)+2) \\
& =f((x-2) f(2)) \cdot f(2) \\
& =0 .
\end{aligned}
$$

Therefore, $f(x)=0$;
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 $0 \leqslant x<2$.
Thus
$$
\begin{array}{l}
f(y f(x))=0, \\
y f(x) \geqslant 2, \\
y \geqslant \frac{2}{f(x)} .
\end{array}
$$

This means, 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
$$
f(x)=\left\{\begin{array}{lc}
\frac{2}{2-x}, & 0 \leqslant x<2 \text { when; } \\
0, & x \geqslant 2 \text { when. }
\end{array}\right.
$$