F8 (24-1, UK) Find all functions $f(x)$ defined on the set of positive real numbers and taking values in the same set that satisfy the conditions:
(1) $f(x f(y))=y f(x)$,
(2) as $x \rightarrow \infty$, $f(x) \rightarrow 0$.
First, we prove that 1 is in the range. For any $x_{0}>1$, let $y_{0}=\frac{1}{f\left(x_{0}\right)}$, substituting into (1) we get
$$
f\left(x_{0} f\left(y_{0}\right)\right)=y_{0} f\left(x_{0}\right)=1,
$$

so 1 is in the range of $f$.
Next, let $y_{0}$ satisfy $f\left(y_{0}\right)=1$, then from (1) we get
$$
f(x)=y_{0} f(x),
$$

so $y_{0}=1$, i.e., $f(1)=1$.
For any positive real number $x$, let $x f(x)=a$, then taking $y=x$ in (1), we get $f(a)=a$.
If $a>1$, then
$$
f\left(a^{2}\right)=f(a f(a))=a f(a)=a^{2},
$$

and by mathematical induction, it is easy to see that
$$
f\left(a^{2^{n}}\right)=a^{2^{n}} \rightarrow+\infty(n \rightarrow \infty),
$$

which contradicts (2).
If $a<1$, by the above discussion (replacing $\frac{1}{a}$ with $a$), this is also impossible.
Therefore, $a=1$, i.e., $f(x)=\frac{1}{x}$.
It is easy to verify that $f(x)=\frac{1}{x}$ satisfies the conditions.