Problem
Find all functions $f:\mathbb{R^+}\rightarrow \mathbb{R^+}$ such that: $$f(1+xf(y))=yf(x+y)$$ for all $x,y \in \mathbb{R^+}$
Progress
I can only prove $f$ is a surjective function. I tried to prove $f$ is an injective function, but I can't. Do you have any ideas?
Let $y=1$. Then $f(1+xf(1))=f(1+x)$, so either $f(1)=1$, or $f(1+e^x)$ is periodic (this part is explained in more detail later). Let's assume the first to start. If $f$ is continuous on $(0,\infty)$ (this constraint is relaxed later), then we can let $x\to0$ so that we get $f(1)=yf(y)=1$ whence $f(x)=1/x$. Checking this solution, we get $$f(1+xf(y))=yf(y+x)\Rightarrow \frac1{1+x/y}=\frac y{y+x},$$ which is true, so $f(x)=1/x$ is a solution.
If we fix some $y\ne1$ such that $f(y)\ne1$ (which exists, because $f(x)=1$ is not a solution) and choose $x$ so that $1+xf(y)=y+x$, then $$x=\frac{y-1}{f(y)-1}\Rightarrow f\Big(\!\frac{yf(y)-1}{f(y)-1}\!\Big)=yf\Big(\!\frac{yf(y)-1}{f(y)-1}\!\Big)\Rightarrow f\Big(\!\frac{yf(y)-1}{f(y)-1}\!\Big)=0\notin\mathbb R^+,$$ which is a contradiction, so either $x=\frac{y-1}{f(y)-1}\le0$ or $y+x=\frac{yf(y)-1}{f(y)-1}\le0$. The second condition implies the first, which is equivalent to $$f(y)\ge1\mbox{ if }y\le1\mbox{ and }f(y)\le1\mbox{ if }y\ge1.$$
If we suppose $a:=\max(f(1),f(1)^{-1})\ne1$, then $f(1+ax)=f(1+x)$ (setting $y=1$). Moreover, this generalizes to $$yf(x+y)=f(1+xf(y))=f(1+axf(y))=yf(ax+y)\Rightarrow f(y+x)=f(y+ax),$$ so letting $y=1-x$, we get $a=f((a-1)x+1)$, so $f(x)=f(1)$ on $[1,a)$ (assuming $a>1$), and by $f(1+ax)=f(1+x)$, it is also constant on $[1,a^2)$, $[1,a^3)$, etc. so that (by induction) it is constant for all $x\ge1$. But then, choosing $y=2$, $f(1+xf(2))=2f(x+2)\Rightarrow$ $f(1)=2f(1)$, since both arguments are greater than $1$, and this is a contradiction. Thus, $f(1)=1$.
If we assume $f$ is continuous at $1$, then if we let $x\to0$, $$\lim_{x\to0}f(y+x)=\lim_{x\to0}\frac{f(1+xf(y))}y=\frac{f(1)}y=\frac1y$$ which implies that $f$ is continuous almost everywhere (discontinuous on a nowhere-dense set) and equals $1/x$ where it is continuous. For the remaining points, let $x$ be small enough that $f(1+xf(y))=\frac1{1+xf(y)}$ and $f(y+x)=\frac1{y+x}$. Then $\frac1{1+xf(y)}=\frac y{y+x}$ so that $f(y)=1/y$ on $(0,\infty)$.
In summary: $f(x)=1/x$ is a solution, and $\operatorname{sgn}(y-1)=\operatorname{sgn}(1-f(y))$. Other than that, it is known that any other solution will have to be discontinuous at $1$. More work to be done, I think...