6- 164 Find all functions $f$ defined on the set of positive real numbers, taking positive real values, and satisfying the conditions:
(1) For all positive numbers $x, y, f(x f(y))=y f(x)$;
(2) As $x \rightarrow \infty$, $f(x) \rightarrow 0$.
[Solution] In (1), take $y=x$, we get
$$f(x f(x))=x f(x)$$

For any $x$, let $x f(x)=a$,
(1)

then (1) becomes $f(a)=a$
(2)
(3)
In (1), take $x=y=a$, we get
$$f(a f(a))=a f(a)$$

Substitute (3) into it, we get $f\left(a^{2}\right)=a^{2}$;
By induction, for all positive integers $n$,
$$f\left(a^{n}\right)=a^{n}$$

In (1), take $x=1, y=a$, we get
$$f(f(a))=a f(1)$$

Substitute (3) into it, we get $a=f(a)=a f(1)$;
By (2), $x$ and $f(x)$ are positive by definition, so $a$ is also positive, divide both sides of the above equation by $a$ to get
$$1=f(1)$$

In (1), take $x=a^{-1}, y=a$, we get
$$f\left(a^{-1} f(a)\right)=a f\left(a^{-1}\right)$$

Substitute (3) into it, we get $f\left(a^{-1} a\right)=a f\left(a^{-1}\right)$,
Substitute (5) into it, we get $f(1)=1=a f\left(a^{-1}\right)$,
$$f\left(a^{-1}\right)=a^{-1}$$

By induction, for all positive integers $n$,
$$f\left(a^{-n}\right)=a^{-n}$$

Now consider the value of $a$ using the given condition (2).
If $a>1$, then by (4) when $n \rightarrow \infty$,
$a^{n} \rightarrow \infty, f\left(a^{n}\right) \rightarrow \infty$, which contradicts (2);
If $a<1$, then by (6) when $n \rightarrow \infty$,
$$a^{-n} \rightarrow \infty, f\left(a^{-n}\right) \rightarrow \infty \text{, which contradicts (2). }$$

Therefore, $a=1$. By the arbitrariness of $x$ in (2), we know that for all $x$,
$$x f(x)=1 \text{ i.e., } f(x)=1 / x \text{.}$$

Upon verification, $f(x)=1 / x$ indeed satisfies conditions (1) and (2).