jmc

From the given equation, $$f\left(f(x)+\frac{1}{x}\right)=\frac{1}{f(x)}.$$Since $y=f(x)+\frac{1}{x}>0$ is in the domain of $f$, we have that $$f\left(f(x)+\frac{1}{x}\right)\cdot f\left(f\left(f(x)+\frac{1}{x}\right)+\frac{1}{f(x)+\frac{1}{x}}\right)=1.$$Substituting $f\left(f(x)+\frac{1}{x}\right)=\frac{1}{f(x)}$ into the above equation yields $$\frac{1}{f(x)}\cdot f\left(\frac{1}{f(x)}+\frac{1}{f(x)+\frac{1}{x}}\right)=1,$$so that $$f\left(\frac{1}{f(x)}+\frac{1}{f(x)+\frac{1}{x}}\right)=f(x).$$Since $f$ is strictly increasing, it must be 1 to 1. In other words, if $f(a)=f(b)$, then $a=b$. Applying this to the above equation gives $$\frac{1}{f(x)}+\frac{1}{f(x)+\frac{1}{x}}=x.$$Solving yields that $$f(x)=\frac{1\pm \sqrt{5}}{2x}.$$Now, if for some $x$ in the domain of $f$, $$f(x)=\frac{1+\sqrt{5}}{2x},$$then $$f(x+1)=\frac{1\pm \sqrt{5}}{2x+2}<\frac{1+\sqrt{5}}{2x}=f(x).$$This contradicts the strictly increasing nature of $f$, since $x<x+1$. Therefore, $$f(x)=\frac{1-\sqrt{5}}{2x}$$for all $x>0$. Plugging in $x=1$ yields $$f(1)=\boxed{\frac{1-\sqrt{5}}{2}}.$$