9. Given that the increasing function $f(x)$ is defined on $(0,+\infty)$, and for any positive number $x$ it satisfies $f(x)$ $f\left[f(x)+\frac{1}{x}\right]=1$, find the value of $f(1)$.
9. Solution: From $f(x) \cdot f\left[f(x)+\frac{1}{x}\right]=1$, substituting $x=1$ gives $f(1) \cdot f[f(1)+1]=1$, thus $f(1)+1>0$.

Let $f(1)=a$, then we have $a \cdot f(a+1)=1, f(a+1)=\frac{1}{a}$.
Substituting $x=a+1$ gives $f(a+1) \cdot f\left[f(a+1)+\frac{1}{a+1}\right]=1, f\left[f(a+1)+\frac{1}{a+1}\right]=\frac{1}{f(a+1)}$.
From (1) and (2), we get $f\left[f(a+1)+\frac{1}{a+1}\right]=a=f(1)$.
Since $f(x)$ is an increasing function on $(0,+\infty)$, we have $f(a+1)+\frac{1}{a+1}=1$, i.e., $f(a+1)=\frac{a}{a+1}$.
From (1) and (3), we get $\frac{1}{a}=\frac{a}{a+1}, a^{2}-a-1=0$, thus $a=\frac{1-\sqrt{5}}{2}$ or $a=\frac{1+\sqrt{5}}{2}$.
If $a=\frac{1+\sqrt{5}}{2}>0$, then $f(a+1)>f(1)$, i.e., $\frac{1}{a}>a, 0<a<1$, which is a contradiction, so we discard it;
If $a=\frac{1-\sqrt{5}}{2}<0$, then it satisfies $f(a+1)<f(1)$. Therefore, $f(1)=\frac{1-\sqrt{5}}{2}$.