Find all functions $f:\mathbb{R}^+\to \mathbb{R}^+$ such that for all $x,y\in\mathbb{R}^+$, $f(x)f(yf(x))=f(x+y)$

Question

Find all functions $f:\mathbb{R}^+\to \mathbb{R}^+$ such that for all $x,y\in\mathbb{R}^+$$$f(x)f(yf(x))=f(x+y)$$

A start: set y=0 to get $f(x)f(0)=f(x)$. So $f(0)=1$ unless $f$ is identically zero.

Answer 1

OK, posting this as an answer since it's the most I can figure out at the moment:

If we suppose that $f$ is continuous and furthermore that it has $1$ in its image, then it is constant (and hence $1$ everywhere).

Proof: Suppose $f(t)=1$. Then applying the functional equation, we find that $f(x)=f(t+x)=f(x+t)=f(x)f(tf(x))$. From this we conclude two things: Firstly, $t$ is a period of $f$. Secondly, cancelling the occurences of $f(x)$ (since, after all, $f$ is never $0$), $f(tf(x))=1$, that is to say, $tf(x)$ also has this same property as $t$, and hence is also a period of $f$.

So if $f$ has any irrational numbers in its image, then it would have two periods with irrational ratio, and hence have arbitrarily small periods; by the continuity assumption, this means $f$ is constant.

But on the other hand, if $f$ has no irrational numbers in its image, then by continuity, this also means $f$ is constant.

(Note that one can also get arbitrarily small periods without using continuity if one assumes that $f(x)<1$ for some $x$; but if you don't assume continuity, I don't know how to show $f$ is constant.)