Example 10 Find all functions defined on the set of positive real numbers, which take positive real values, and satisfy the conditions:
(1) For all positive numbers $x, y, f(x f(y))=y f(x)$;
(2) As $x \rightarrow \infty$, $f(x) \rightarrow 0$.
(24th IMO)
Solution: In (1), let $y=\frac{1}{f(x)}$, we get
$$
f(x f(y))=1 \text {. }
$$

This shows that 1 is a value that $f$ can take.
Now, let $y$ be such that $f(y)=1$. From (1), we get
$$
f(x)=y f(x) \text {. }
$$

Thus, $y=1$. Therefore, $f(1)=1$.
For any $x>0$, let $x f(x)=a$. Taking $y=x$ in (1), we get $f(a)=a$.
If $a>1$, then
$$
f\left(a^{2}\right)=f(a f(a))=a f(a)=a^{2} \text {. }
$$

By mathematical induction, it is easy to prove
$$
f\left(a^{2^{n}}\right)=a^{2^{n}} \rightarrow+\infty \text {, }
$$

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