Solving the functional equation $(x+1)f\left(\frac{y}{f(x)}\right)=f(x+y)$

Question

Solve functional equation:

Find all strictly monotone functions $f:(0,+\infty)\to(0,+\infty)$ such that $$(x+1)f\left(\frac{y}{f(x)}\right)=f(x+y),\forall x,y>0\text.$$

Answer 1

This is just something that is missing from David's answer. (Which i find very clever)
We have to prove first that $f_{0^+}$ is a real number. (Check line 5 at David's answer )

We need to prove first that $f$ is strictly increasing.
Substitute $x=1,y=f(1)$ in the equation. (Both are positive, so we have no problem)

We have $$2\cdot f(1)=f(1+f(1))\Rightarrow f(1)=\frac{f(1+f(1))}{2}<f(1+f(1))$$ since $f(1+f(1))>0$.

But $1<1+f(1)$ and $f(1)<f(1+f(1))$
Knowing that $f$ is strictly monotone, we see that $f$ is strictly increasing and since it's domain is $(0,+\infty)$:
$$\lim_{y \to 0^{+}}f(y)=f_{0^+}\in \mathbb R$$Combining this and David's approach we get the desired result.