Example 10 Suppose the function for all $x>0$ has $f(x)>-\frac{1}{x}$ and is defined for all $x>0$, and satisfies:
(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 $f(1)$.
Solution: Let $f(1)=a$, then by condition (3) taking $x=1$, we have $a f(a+1)=1$, i.e., $f(a+1)=\frac{1}{a}$.

Again, in (3) taking $x=a+1$, we have $f(a+1)$ $f\left[f(a+1)+\frac{1}{a+1}\right]$, i.e.,
$f\left(\frac{1}{a}+\frac{1}{a+1}\right)=a=f(1)$, by condition (1) we have $\frac{1}{a}$ $+\frac{1}{a+1}=1$, solving this gives $a=\frac{1 \pm \sqrt{5}}{2}$.

If $a=\frac{1+\sqrt{5}}{2}$, then $1<a=f(1)<f(1+a)=$ $\frac{1}{a}<1$, a contradiction.
$$
\therefore f(1)=a=\frac{1-\sqrt{5}}{2} \text {. }
$$