Find all functions $f:\mathbb{R}^+\to \mathbb{R}^+$ such that for all $x,y\in\mathbb{R}^+$, $f(x)f(yf(x))=f(x+y)$ 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.
 A: Actually we don't need to use the condition that the function is continuous.
Denote the condition $f(x)f(yf(x))=f(x+y)$ by $(1)$
Case I: for some $t$ in $\mathbb{R}^+$, $f(t)>1$.
We put $(x,y)=(t,t/(f(t)-1))$ in $(1)$, and consequently we have $yf(x)=x+y$. Thus we have $f(x)f(yf(x))=f(x+y)$ with $yf(x)=x+y$ and thus $f(yf(x))=f(x+y)>0$. Then $f(t)=f(x)=1$, a contradiction.
Case II: for all $t$ in $\mathbb{R}^+$, $1≥f(t)>0$. Moreover, there is at least one $t$ such that $f(t)=1$.
Then we prove that $f$ is constant. Otherwise, there is another positive real number $r$ such that $f(r)<1$. 
We put $(x,y)=(t,kt)$ where $k=1,2,\dotsc$.
We have that $f(t)f(ktf(t))=f((k+1)t)$, that is, $f(kt)=f((k+1)t)$. Then $f(kt)=f((k-1)t)=\dotsb=f(t)=1$ for any postive integer $k$. 
There is an integer $n$ larger than $r/t$, which means $nt>r$. We put $(x,y)=(r,nt-r)$, and we have $f(r)f((nt-r)f(r))=f(nt)=1$. Thus $f((nt-r)f(r))=1/f(r)>1$, a contradiction. 
Case III: for all $t$ in $\mathbb{R}^+$, $1>f(t)>0$. 
Then we prove that $f$ is injective firstly. Otherwise, we will have that $f(u)=f(v)$ with $u>v>0$. Let $(x,y)=(v,u-v)$ in $(1)$, we have $f(v)>f(v)f((u-v)f(v))=f(u)$ (because $1>f((u-v)f(v))>0$), a contradiction. 
Then, let $x=1$ in $(1)$, we have that (denote $f(1)$ by $c$) $cf(cy)=f(y+1)$. Let $y=1/f(x)$ in $(1)$, we have that $cf(x)=f(x+1/f(x))$. Consider the two equation above, let $y=x/c$, we have that $f(x/c+1)=cf(x)=f(x+1/f(x))$, which, by using that $f$ is injective, means that $x/c+1=x+1/f(x)$ for all $x$.  Then we have that $f(x)=1/(ax+1)$ where $a=1/c-1$ is a postive constant. It is easy to substantiate that all functions in this form satisfy the given equation $(1)$. 
QED.
A: 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.)
