6・160 (1) Let the function $f(x)$ be defined for all $x>0$, and satisfy the following conditions:
(1) The function $f(x)$ is strictly increasing on $(0,+\infty)$;
(2) For all $x>0$, $f(x)>-\frac{1}{x}$;
(3) For all $x>0$, $f(x) f\left(f(x)+\frac{1}{x}\right)=1$. Find the function value $f(1)$.
(2) Provide a function that satisfies the three conditions in (1).


[Solution] (1) Let $f(1)=a$, then by condition (3), when $x=1$, we have $a f(a+$
1) $=1$. That is, $f(a+1)=\frac{1}{a}$.

Now let $x=a+1$, then by condition (3) we get
$$f(a+1) f\left(f(a+1)+\frac{1}{a+1}\right)=1$$

That is, $f\left(\frac{1}{a}+\frac{1}{a+1}\right)=a$,
Therefore, $\quad f\left(\frac{1}{a}+\frac{1}{a+1}\right)=f(1)$.
By condition (1), we get $\frac{1}{a}+\frac{1}{a+1}=1$,
Solving this, we get
$$a=\frac{1 \pm \sqrt{5}}{2}$$

If
$$a=\frac{1+\sqrt{5}}{2}$$

then $\square$
$$10$$

That is, $f\left(x_{2}\right)>f\left(x_{1}\right)$. Therefore, $f(x)$ is strictly increasing on $(0,+\infty)$, which means $f(x)$ satisfies condition (1); Since
$$f(x)+\frac{1}{x}=\frac{3-\sqrt{5}}{2 x}$$

$f(x)$ satisfies condition (2); Finally, since
$$f(x) f\left(f(x)+\frac{1}{x}\right)=\frac{a}{x} \cdot \frac{a}{\frac{a}{x}+\frac{1}{x}}=\frac{a^{2}}{a+1}=1$$

$f(x)$ satisfies condition (3). Therefore, the function
$$f(x)=\frac{a}{x}$$

meets all three conditions in the problem.