Finding all $f:\mathbb{R}^+\to \mathbb{R}^+$ satisfying $f(x)\cdot f\big(yf(x)\big)=f\big(y+f(x)\big)$ for all $x,y \in \mathbb{R}^+$ Find all $f:\mathbb{R}^+\to \mathbb{R}^+$ that satisfy
$$f(x) f\big( y f(x) \big) = f\big( y + f(x) \big)$$
$\forall x,y \in \mathbb{R}^+$.
 A: The following is a complete proof that $f\equiv 1$ is the only solution.
Suppose $u,v\in f(\mathbb R^+)$. Then $u = f(x)$ for some $x\in\mathbb R^+$. Therefore the functional equation tells you that $f(x) f(vf(x)) = f(v+f(x))$ or in other words $$u f(uv)=f(u+v).$$ (Note that we have used the fact that $f(x)\in\mathbb R^+$ for $x\in\mathbb R^+$, so that the functional equation holds for $u$ and $v$.)
But as we only assumed that $u,v\in f(\mathbb R^+)$, the same also holds if we interchange the roles of $u$ and $v$, therefore $$v f(uv)=f(u+v)$$ must also hold. Combining the two equations, we have $u f(uv)=v f(uv)$, but because $f(x)$ is positive for all $x\in\mathbb R^+$, we can divide by $f(uv)$. This gives us $u=v$.
What does this mean? We assumed $u,v\in f(\mathbb R^+)$ and proved that $u=v$. This means that $f(\mathbb R^+)$ only has a single element. In other words, $f$ is a constant function, i.e. $f(x)=c$ for all $x$, where $c$ is some positive constant. But from the functional equation we now have that $c$ must satisfy $c^2 = c$, and since $c$ is positive, this is only possible if $c=1$. This completes the proof.
A: Assume there exists an $x_0$ such that $f(x_0)>1$. Then with $y:=\frac{f(x_0)}{f(x_0)-1}$ (i.e. the solution of $yf(x_0)=y+f(x_0)$) we obtain the contradiction $f(x_0)=\frac{f(y+f(x_0))}{f(yf(x_0))}=1$.
Therefore, $f(x)\le 1$ for all $x$.
Assume $q:=f(x_1)<1$. 
Then define recursively $x_{n+1}=\frac{x_n}q+q$ to find 
$$f(x_{n+1})=f\left(\frac{x_n}{q}+f(x_1)\right)=f(x_1)f\left(\frac{x_n}qf(x_1)\right)=qf(x_n),$$ hence by induction $f(x_n)=q^n$.
Select $n\in\mathbb N$ with $q^n<f(1)$.
Then with $y:=1-q^n$ we find 
$$f(1)=f(y+f(x_n))=f(x_n)f(yf(x_n))\le f(x_n)=q^n,$$
contradiction. Therefore the assumption $f(x_1)<1$ cannot hold. We conclude $f(x)=1$ for all $x$.
