7. Let $f(x)$ be a monotonically increasing function defined on $(0,+\infty)$, and for any positive number $x$, we have
$$
f\left(f(x)+\frac{1}{x}\right)=\frac{1}{f(x)} .
$$

Then $f(1)=$ $\qquad$
7. $\frac{1-\sqrt{5}}{2}$.

In equation (1), let $x=1$, we get
$$
f(f(1)+1)=\frac{1}{f(1)} \text {. }
$$

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

Since $f(x)$ is strictly increasing on $(0,+\infty)$, we have $\frac{1}{a}+\frac{1}{a+1}=1$.
Solving this, we get $a=\frac{1 \pm \sqrt{5}}{2}$.
When $a=\frac{1+\sqrt{5}}{2}$, $1+a>1$.
But $f(a+1)=\frac{1}{a}<f(1)=a$, which contradicts the given condition;
When $a=\frac{1-\sqrt{5}}{2}$, $1+a<1$.
In this case, $\frac{1}{a}<a$, satisfying $f(a+1)<f(1)$.
Therefore, $f(1)=\frac{1-\sqrt{5}}{2}$.