Easy functional equation: $ f \big( 2 f ( x ) + f ( y ) \big) = 2 x + f ( y ) $ 
Find all functions $ f : \mathbb R \to \mathbb R $ such that:
$$ f \big( 2 f ( x ) + f ( y ) \big) = 2 x + f ( y ) \qquad \forall x , y \in \mathbb R \text . $$

If you put $ x = y = 0 $, you get $ f \big( 3 f ( 0 ) \big) = f ( 0 ) $. What deductions about $ f ( 0 ) $ can you then make?
Clearly from above $ f ( 0 ) = 0 $ is a solution... so,
putting $ x = 0 $ gives $ f \big( 2 f ( 0 ) + f ( y ) \big) = f  ( y ) $.
$ \implies f \big( f ( y ) \big) = f ( y ) $.
So $ f ( x ) = x $ is a solution, but is it the only one?
I think it probably is, but how to prove?
 A: $$f(2f(x)+f(y))=2x+f(y)\qquad \forall x,y \in \mathbb{R}.$$
Interchaning $x$ and $y$ you get
$$f(f(x)+2f(y))=f(x)+2y \,.$$ 
Claim 1: $f(x)$ is 1 to 1.
Indeed, if $f(x)=f(y)$ then
$$2x+f(x)=2x+f(y)=f(2f(x)+f(y))=f(f(x)+2f(y))=f(x)+2y $$
This implies that $x=y$.
Now, you can do part of what you did:
$$f(2f(0)+f(y))=f(y)\qquad \forall y \in \mathbb{R}.$$
Since $f$ is 1 to 1 you get
$$2f(0)+f(y)=y \,.$$
Thus
$$f(y)=y-2f(0)\,.$$
Setting $y=0$ you get $f(0)=0$ and thus $f(x)=x$ is the only solution.
A: $$
f(2f(x)+f(y)) = 2x+f(y)
$$
Now consider values of $x$ and $y$ such that $2f(x)=-f(y)$. Now, we can write that
$$
f(0) = 2(x-f(x))
$$
Therefore, we may write that
$$
f(x) = x-\frac{f(0)}{2}
$$
Substituting $x=0$, we get
$$
f(0) = -\frac{f(0)}{2}
$$
And therefore $f(0)=0$, thereby giving $f(x)=x$.
This analysis required an assumption: that values of $x$ and $y$ could be found such that $2f(x)=-f(y)$. This can be confirmed - for any value of $y$, we have that
$$
f(2f(x)+f(y))=2x+f(y)
$$
Therefore, for any given real number, $a$, we may choose $x_a=\frac{f(y)-a}{2}$, and thus we have $f(2f(x_a)+f(y))=a$, and thus the function may take the value $a$. As this must be true of any $a$, we know that the function must take all values, and thus there must exist a solution to $2f(x)=-f(y)$ for any $x$ or for any $y$.
