Finding $f:\mathbb R\to\mathbb R$ satisfying $f\bigl(yf(x+y)+f(x)\bigr)=4x+2yf(x+y)$ 
Find $f:\mathbb R\to\mathbb R$ satisfying $f\bigl(yf(x+y)+f(x)\bigr)=4x+2yf(x+y)$.

This is my attempt:
\begin{align}
& \text{Let } f(0)=t \text. \\
&P(0, 0): f(t)=0 \text. \\
&P(0, t): t = 0 \text. \implies f(0)=0 \text. \\
\\
&\text{Assume) } \exists \ t \text{ s.t. } t \ne 0 \text{ and } f(t)=0 \text. \\
&P(t, 0): f(0)=4t \text. \\
&P(0, 0): f(4t)=0 \text. \\
&P(4t, 0): f(0)=16t \text. \\
&\implies 4t=16t \text, \ t=0 \text. \\
&\therefore f(t)=0 \iff t=0 \text. \\
\\
&\text{Let } f(a)=f(b) \text. \\
&P(a, 0):f\bigl(f(a)\bigr)=4a \text. \\
&P(b, 0):f\bigl(f(b)\bigr)=4b \text. \\
&\implies a = b \text. \\
&\therefore f(a)=f(b) \iff a = b \text.
\end{align}
My expectation is the $f(x)=2x$.
Can you help me getting the perfect process of finding the function $f$?
Edit:
I found more info...
\begin{align}
& P(x, x): f\bigl(xf(2x)+f(x)\bigr)=4x+2xf(2x) \text. \\
& P(x, -x): f\bigl(f(x)\bigr)=4x \text. \\
& P(0, x): f\bigl(xf(x)\bigr)=2xf(x) \text. \\
& x=1 \text; \ f\bigl(f(1)\bigr)=2f(1) \text. \\
& \text{From } P(1, -1): f\bigl(f(1)\bigr)=4 \text. \\
& \therefore f(1)=2 \text.
\end{align}
 A: You can prove that the only $ f : \mathbb R \to \mathbb R $ satisfying
$$ f \bigl ( y f ( x + y ) + f ( x ) \bigr ) = 4 x + 2 y f ( x + y ) \tag 0 \label 0 $$
for all $ x , y \in \mathbb R $ is the doubling function $ f ( x ) = 2 x $, even without further assumptions (like continuity of $ f $). It's straightforward to verify that the doubling function is indeed a solution. To prove the converse, put $ y = 0 $ in \eqref{0} to get
$$ f \bigl ( f ( x ) \bigr ) = 4 x \tag 1 \label 1 $$
for all $ x \in \mathbb R $. In particular, \eqref{1} shows that $ f $ is injective. Letting $ a = f \left ( \frac 1 2 \right ) $, \eqref{1} gives $ f ( a ) = 2 $. Setting $ y = a - x $ in \eqref{0} and using \eqref{1} you get
$$ f \bigl ( 2 ( a - x ) + f ( x ) \bigr ) = 4 x + 4 ( a - x ) = 4 a = f \bigl ( f ( a ) \bigr ) = f ( 2 ) \tag 2 \label 2 $$
for all $ x \in \mathbb R $. By injectivity of $ f $, \eqref{2} shows
$$ 2 ( a - x ) + f ( x ) = 2 \text , $$
or equivalently
$$ f ( x ) = 2 x + b $$
for all $ x \in \mathbb R $, where $ b = 2 - 2 a $. Therefore,
$$ f \bigl ( f ( x ) \bigr ) = 2 ( 2 x + b ) + b = 4 x + 3 b $$
for all $ x \in \mathbb R $, which together with \eqref{1} yields $ b = 0 $. Hence, $ f ( x ) = 2 x $ for all $ x \in \mathbb R $, as desired.
A: $$f(yf(x+y)+f(x))=4x+2yf(x+y) $$
Generally, just look at the function and try to see something that will simplify or try random useful things.
One useful thing, generally, is to set $y = x$ to reduce to a single function, since this is a single variable function we can look at it along a common value and potentially find a function that works:
$$f(xf(2x)+f(x))=4x+2xf(2x) $$
What jumps out at me immediately is that if $f(x) = x$ it almost works. We really need $f(x) = 2x$ though to capture the additional factor. Hence
$$2(x\cdot2(2x) + 2x) = 4x + 2x\cdot2(2x))$$
$$8x^2 + 4x = 4x + 8x^2$$
And so it works.
I'll leave the rest to you. It's not the perfect process, depending on how you define perfect. This does not imply that $f(x) = 2x$ is the only solution but it is a particular solution from which you can work to build off of.
Most of the time these types of problems are relatively simple and obvious because these functional relationships can't be arbitrary and the more terms and relationships you have the harder it is to satisfy.
A: $$f: \mathbb{R}\to\mathbb{R}, f(yf(x+y)+f(x))=4x+2yf(x+y)$$
$$\text{let } f(0)=t$$
$$P(0, 0): f(t)=0 $$
$$P(t, 0): f(0)=4t=t \Rightarrow t=0 $$
$$\therefore f(0)=0$$
$$\text{let } f(a)=f(b) $$
$$P(a, 0):f(f(a))=4a $$
$$P(b, 0):f(f(b))=4b $$
$$\therefore f(a)=f(b)\Rightarrow a=b$$
$$P(x,0): f(f(x))=4x $$
$$P(0, x): f(xf(x))=2xf(x)$$
$$f(f(1))=4, f(1f(1))=2f(1)\Rightarrow f(1)=2$$
$$f(f(1/2))=2, f(1)=2\Rightarrow f(1/2)=1$$
$$A=\{x|f(x)= 2x,x\in\mathbb{R}\}$$
$$x\in A\Rightarrow f(f(x))=4x= 2f(x)\Rightarrow f(x)=2x\in A$$
$$x\in A \Rightarrow f(f(x/2))=2x=f(x) \Rightarrow f(x/2)=x\Rightarrow x/2 \in A$$
$$\therefore x\in A \Rightarrow 2^k x\in A, k \in \mathbb{Z}$$
$$P(x-1/2,1-x):f(1-x+f(x-1/2))=2x$$
$$x\in A \Rightarrow f(1-x+f(x-1/2))=f(x)\Rightarrow 1-x+f(x-1/2)=x\Rightarrow f(x-1/2)=2x-1$$
$$\therefore x\in A \Rightarrow x-k/2\in A,k\in \mathbb{N}$$
$$1\in A \Rightarrow 2^n \in A \Rightarrow 2^n-m\in A, n,m \in \mathbb{N}\Rightarrow \mathbb{Z}\subset A$$
$$m\in A\Rightarrow m\ 2^{-n}\in A, m\in \mathbb{Z}, n \in \mathbb{N}$$
Any real number can be approximated as $m\ 2^{-n}$ with any accuracy, so if $f$ is continuous, then $A=\mathbb{R}$.
