Example 7 Let the function $f(x)$ be defined for all $x>0$ and satisfy:
(i) $f(x)$ is strictly increasing on the interval $(0,+\infty)$;
(ii) For all $x>0$, $f(x)>-\frac{1}{x}$;
(iii) For all $x>0$, $f(x) f\left(f(x)+\frac{1}{x}\right)=1$.
(1) Find $f(1)$.
(2) Find a function $f(x)$ that satisfies the above conditions.
(1) Let $f(1)=a$. Then by (iii), setting $x=1$, we get
$$
a f(a+1)=1 \Rightarrow f(a+1)=\frac{1}{a} \text {. }
$$

Now let $x=a+1$, we have
$$
\begin{array}{l}
f(a+1) f\left(f(a+1)+\frac{1}{a+1}\right)=1 \\
\Rightarrow f\left(\frac{1}{a}+\frac{1}{a+1}\right)=a \\
\Rightarrow f\left(\frac{1}{a}+\frac{1}{a+1}\right)=f(1) .
\end{array}
$$

By the strictly increasing nature of the function $f$, we have
$$
\frac{1}{a}+\frac{1}{a+1}=1 \Rightarrow a=\frac{1 \pm \sqrt{5}}{2} \text {. }
$$

If $a=\frac{1+\sqrt{5}}{2}$, then
$$
1<a=f(1)<f(1+a)=\frac{1}{a}<1,
$$

which is a contradiction.
Therefore, the only possibility is $a=\frac{1-\sqrt{5}}{2}$.
$$
\text { (2) } f(x)=\frac{1-\sqrt{5}}{2 x} \text {. }
$$