Obviously $f(x)\neq 0$ always. The equation is $f\left(f(x)+\frac{1}{x}\right)=\frac{1}{f(x)}$. Now put $f(x)+\frac{1}{x}$ instead of $x$, we get \[f\left(\frac{1}{f(x)}+\frac{1}{f(x)+\frac{1}{x}}\right)=f(x)\]The function is strictly increasing and hence injective. This implies \[\frac{1}{f(x)}+\frac{1}{f(x)+\frac{1}{x}}=x\]If $x=1$, we get $f(1)=\frac{1\pm \sqrt{5}}{2}$. We cannot have $f(1)=\frac{1+\sqrt{5}}{2}$, because $f(1+f(1))>f(1)$, and $f(1+f(1))f(1)=1$. Edit: By the problem condition, $f(x)+\frac{1}{x}$ is always positive (or else $f$ of it won't be defined). Thus it's okay to put it as $x$.