Find a function $f:\Bbb R^+\to\Bbb R$ that is strictly increasing and satisfies:
$f(x) > -\frac1x$ for all $x > 0$
$f(x)f\left(f(x)+\frac1x\right) = 1$ for all $x > 0$
Attempt:
If $f(x)+\frac1x= g(x),$ then $f(x) = \frac1{f(g(x))}.$
Replace $x$ with $g(x).$ Then, we get
$$f(g(x)) = \frac1{f(g(g(x)))}$$
and so, $$f(x) = f(g(g(x)))$$
If we assume $f$ is injective, this means $$\begin{aligned}g(g(x)) = x\\f(g(x)) + 1/g(x) = x\\\frac1{f(x)} + \frac1{f(x) + 1/x} = x.\end{aligned}$$
This leads to a quadratic in $f(x)$ which finally gives
$$f(x) = \frac{1 + \sqrt{1 + 4/x}}2$$
Now, to complete this, I need to find non - injective functions $f$ which also satisfy this (or prove that none exist). Since I didn't use the first condition at all, I think there are some non-injective solutions, but I cannot find them.