find $f:\mathbb{R}\rightarrow\mathbb{R}$ that satisfies $\displaystyle f(yf(x+y)+f(x))=4x+2yf(x+y)$
it's clear that $f(f(x))=4x$ notice that $f$ is injective $\rightarrow f(4x)=4f(x)$ therefore, $f(0)=0$
and we see that $f(xf(x))=2xf(x)\therefore 4=f(f(1))=2f(1)\rightarrow f(1)=2$
consider that $f(2)=f(f(1))=4$ And then we replace $P(x,1-x)$ we get
$f\Big(2(1-x)+f(x)\Big)=f\Big((1-x)f(1)+f(x)\Big)=4x+2(1-x)f(1)=4x+2(1-x)2=4=f(2)$
since $f$ is injective so $2-2x+f(x)=2\Longrightarrow f(x)=2x$