6. Given that the function $f(x)$ is defined and increasing on $(0,+\infty)$, and satisfies $f(x) \cdot f\left(f(x)+\frac{1}{x}\right)=1$. Then $f(1)=$ ).
(A) 1
(B) 0
(C) $\frac{1+\sqrt{5}}{2}$
(D) $\frac{1-\sqrt{5}}{2}$
6. (D).

Let $f(1)=a \cdot x=1$. From the given functional equation, we have
$$
\begin{array}{l}
f(1) f(f(1)+1)=1, \\
\quad a \int(a+1)=1, f(a+1)=\frac{1}{a} .
\end{array}
$$

Let $x=a+1$. Then
$$
\begin{array}{l}
f(a+1) f\left(f(a+1)+\frac{1}{a+1}\right)=1, \\
\frac{1}{a} f\left(\frac{1}{a}+\frac{1}{a+1}\right)=1, \\
f\left(\frac{1}{a}+\frac{1}{a+1}\right)=a=f(1) .
\end{array}
$$

Since $f(x)$ is an increasing function, we have $\frac{1}{a}+\frac{1}{a+1}=1$, solving which gives
$$
a=\frac{1 \pm \sqrt{5}}{2} .
$$

When $f(1)=\frac{1+\sqrt{5}}{2}$, we have
$$
1<a=f(1)<f(a+1)=\frac{1}{a}<1
$$

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