Find all strictly monotone $f:(0,+\infty) \to (0, +\infty)$ such that $f(\frac{x^2}{f(x)})=x.$ Find all strictly monotone functions $f:(0,+\infty) \to (0,+\infty)$ such that
$$f\left(\frac{x^2}{f(x)}\right)=x.$$
My try: it is clear that $f$ is surjective. And because it is monotone it must also be injective. Therefore we can take $f^{-1}$ from both sides: $x^2 = f(x) \cdot f^{-1}(x)$.
We can take $x = f(y)$ (because of surjectivity) and get that: $\frac{f(y)}{y} = \frac{f(f(y))}{f(y)}$.
So, if we define $g(x) = \frac{f(x)}{x}$ we have that $g(y) = g\big(f(y)\big)$ and I was hoping to prove that $g$ is injective so we would have $f(x) = x$ only. But I couldn't figure that last step. There may be a better way to deal with this problem.
EDIT: There is another solution on AOPS, problem 312.
 A: Consider $h : x \mapsto \ln\big(f(e^x)\big) $.
You need to prove that $h(x)+h^{-1}(x)=2x$ (I leave it to you because it is simple). we have that $h$ is increasing (also easy to prove by contradiction or another way).
Now consider $ n \in \mathbb N$ and define $r_n:= h^n(x)$ and $s_n:=h^{-n}(x)$.
We have:
$$r_{n+1}+r_{n-1}=h(r_n)+h^{-1}(r_n)=2r_n \text,$$
and similarly
$$s_{n-1} +s_{n+1}=2s_n\text.$$
Therefore:
$$ r_n= \lambda(x) + \mu(x)n $$
(and $ s_n= \alpha(x) + \beta(x)n $).
Now, let's prove that $h$ is continuous:
let $x , y \in \mathbb R $ such that $x>y$.
$h(x) - h(y) < h(x) - h(y) + h^{-1}(x) - h^{-1}(y)$ , because $h^{-1}$ is also increasing.
Therefore $h(x) - h(y) < 2(x-y)$ or $|h(x) - h(y) |< 2|x-y|$.
Thus $h$ is continuous.
I wasn't able to proceed from here, but given the continuity you can use the linked post's answer by Martin R.
A: Starting from $\frac{f(f(y))}{f(y)} = \frac{f(y)}{y}$:
For each fixed $y>0$, let $y_n=f^n(y)$, $n\in {\Bbb Z}$ be its orbit. The above condition on $f$ yields:
$$ \forall n \in {\Bbb Z}: \frac{y_{n+1}}{y_n}=\frac{y_n}{y_{n-1}} =: \lambda(y)$$
for some $\lambda(y)>0$ independent of $n$. Thus, $f^n(y) = y \lambda(y)^n$ for all $n\in {\Bbb Z}$.
Monotonicity of $f$, whence of $f^n$, implies:
$$ 0<x<y \Rightarrow \forall n \in {\Bbb Z} : 0 < x \lambda(x)^n < y \lambda(y)^n$$
which can be satisfied iff $\lambda(x)=\lambda(y)$. Thus, $f(x) =\lambda x$ for some $\lambda>0$.
