Partial Solution: Since $ 0 $ lies in the co-domain of $ f $, we have $$ f\left(f\left(y\right)\right) = yf\left(1\right) \xrightarrow[\text{}]{\text{iterate}} f\left(yf\left(1\right)\right) = f\left(y\right)f\left(1\right) $$In particular, $ f\left(0\right) = f\left(0\right)f\left(1\right) $, which means $ f\left(0\right) = 0 $ or $ f\left(1\right) = 1 $. If the latter holds, we get $ f\left(f\left(y\right)\right) = y $ and thus $ f\left(f\left(0\right)\right) = 0 = f\left(2f\left(0\right)\right) $. It follows that $ f\left(0\right) = 0 $ in this case. We conclude that $ f\left(0\right) $ is always zero. Finally, $ 0 = f\left(f\left(x\right)\right) = xf\left(1\right) $, $ f\left(1\right) $ equals to zero as well.

Let $P(x,y)$ denote the assertion. $P(x,0) $ and $P(0,0)$ gives that $f(f(x)+f(0))=0$ and $f(2f(0))=0$ $P(2f(0),y)$ and $P(2f(0),0)$ gives $f(f(y))=yf(1)$ and $f(f(0))=0$ . If $f(1) \neq 0$ we have that $f$ is bijective , and $f(2f(0))=f(f(0))$ implies $f(0)=0$ and from $f(f(x)+f(0))=0$ we have $f(f(x))=0$ , but this contradicts surjectivity. So we must have $f(1)=0$ . $P(1,1)$ gives $f(1)=f(0)=0$ See below for complete solution.

It's not hard to get $f(0)=f(1)=0$. Suppose $f(p)\ne 0$, $p>1$. Let $q=f(p)+f\left(\frac{p-1}{f(p)}\right)$. $P\left(p, \frac{p-1}{f(p)}\right)$ gives us $$f(q)=p-1.$$Comparing $P(q, 0)$ with $P(q, 1)$ gives $$0=f(f(q))=f(p),$$a contradiction. All such numbers are $0$ and all numbers greater than or equal to $1$, because the function $$f(x)=\begin{cases}1\textrm{, if }0<x<1\\0\textrm{ otherwise}\end{cases}$$solves the equation.

Anyone can solve this functional equation which $f:\mathbb{R^+}\rightarrow\mathbb{R^+}$ ?