In a literary work "functional equation", I found a functional equation may be difficult to me, that is, provided that a real differential function on the real line 
I think maybe it needs first to prove the map is surjection or injection? I know the answer is 
But how to prove it strictly?
I can't prove this with complete generality but I can weaken the assumptions on the answer above - we merely assume that $f$ has a zero, not that $f(0)=0$. Hopefully someone else can use the ideas to finish the proof off. Let $f$ be a non-constant solution to the functional equation and set $c:=f(0)$.
Firstly, the given solution will follow if $f$ is injective. Indeed, in this case we have that $x+f(y)=y+f(x)$ for all $x,y\in\mathbb R$, so simply set $y=0$.
Next, assume that there exists $\epsilon>0$ such that $(-\epsilon,\epsilon)\subset f(\mathbb R)$. Differentiating the functional equation with respect to $x$ gives
$$f'(x+f(y))=f'(x)f'(y+f(x)) \qquad (1)$$
and so if $f'(x_0)=0$, $f'(x)=0$ for all $x\in(x_0-\epsilon,x_0+\epsilon)$. It quickly follows that either $f'\equiv0$, contradicting the fact that $f$ is not constant, or $f'(x)\neq0$ for all $x$. Rolle's theorem then implies that $f$ is injective.
Now assume that $f$ has a zero. By continuity there exists $\delta>0$ such that $[0,\delta]\subset f(\mathbb R)$ or $[-\delta,0]\subset f(\mathbb R)$. In either case, it will follow that $(-\epsilon,\epsilon)\subset f(\mathbb R)$ for some $\epsilon>0$. The proofs of each case are almost identical so I will only prove the case where $[0,\delta]\subset f(\mathbb R)$. By $(1)$ it follows that if $f'(x_0)=0$, $f'(x)=0$ for all $x>x_0$. Set $x_1:=\inf\{x\in\mathbb R\,:\,f'(x)=0\}$. If $x_1=\infty$ then $f$ is injective and if $x_1=-\infty$ then $f$ is constant, so we assume $x_1\in\mathbb R$.
If $M:=f(x_1)$ then $M$ is either the global maximum or minimum of $f$. Indeed, $f$ is monotone on $(-\infty,x_1)$ and constant on $[x_1,\infty)$ so $f$ must either increase or decrease to $M$. If $M=\max f$, then $M\ge\delta>0$ and so if $z_0$ is a zero of $f$ then $z_0<x_1$. But if $x<z_0$, $f(x)<f(z_0)=0$, so by continuity $f(\mathbb R)$ contains an interval $(-\epsilon,\epsilon)$.
Suppose now $M=\min f$. Then $M\le0$; if $M<0$ then again we have $(-\epsilon,\epsilon)\subset f(\mathbb R)$, so assume $M=0$. By putting $x=y=0$ in $(1)$ we get $f'(c)=f'(0)f'(c)$. If $c<x_1$, then $f'(c)\neq0$, so $f'(0)=1$. But $f$ is either decreasing or constant, so this is a contradiction. Hence assume $c\ge x_1$. Then $f(c)=0$, and so $$c=f(0+f(c))=f(c+c)=f(2c).$$ But $c=f(0)\ge0$, so $2c\ge c$ and hence $c=f(2c)=0$. But @universalset already provided a solution to the case where $c=f(0)=0$, so we are done. (In fact, the conclusion we reach contradicts the assumption that $x_1<\infty$, but this is not so important.)
Obviously this involves the non-trivial assumption that $f(x)=0$ for some $x\in\mathbb R$. We still need to prove this is the case. Assume for a contradiction that $f>0$ everywhere (the negative case will be analogous). If $\inf f=0$ then essentially the same proof as above implies the existence of $x_0$ such that $f(x)=\max f$ for all $x\ge x_0$ and $f'(x)>0$ for $x<x_0$. However, it is potentially possible that $\lim_{x\to-\infty}f(x)=0$ in which case the argument breaks down. I also have not considered the case where $f(x)\ge\epsilon>0$ for all $x\in\mathbb R$. Perhaps someone else can give it a go. If it helps, one may show that $f'(0)\neq0$ and that $f'(x-c)=1/f'(0)$ whenever $f'(x)\neq0$. I did not use this fact so did not include its proof, but it is straightforward enough using $(1)$ and the fact that $f'$ is not identically zero.
Note: the original version of this answer asserted that if $f(x)$ satisfies the functional equation, then so did $f(x)+c$. This is not obviously true. I'm making this CW in the hope that someone can build on this base and finish the proof appropriately.
Suppose that $f(0) = 0$. In particular, by substituting $y=0$, we obtain $f(x) = f(f(x))$.
Take the derivative with respect to $x$, obtaining \begin{equation}f^\prime(x+f(y)) = f^\prime(x)f^\prime(y+f(x))\qquad (*)\end{equation} for all $x$ and $y$.
Substitute $y = 0$ in $(*)$, obtaining $$f^\prime(x) = f^\prime(x+f(0)) = f^\prime(x)f^\prime(f(x)). \qquad (1)$$
Substitute $x=0$ in $(*)$ and then rename $y$ to $x$, obtaining $$f^\prime(f(x)) = f^\prime(0)f^\prime(x+f(0)) = f^\prime(0)f^\prime(x). \qquad(2)$$
Substituting (1) into (2), $f^\prime(f(x)) = f^\prime(0) f^\prime(x) f^\prime(f(x))$.
So for each $x$, either $f^\prime(f(x)) = 0$ or $f^\prime(0)f^\prime(x) = 1$.
Now if $f^\prime(0) = 0$, this implies that $f^\prime(f(x)) = 0$ for all $x$, and hence by (1) that $f^\prime(x) = 0$ for all $x$, so that $f(x)$ is constant (and thus $f(x) = 0$ since $f(0)=0$).
Otherwise, setting $x = 0$ gives $f^\prime(0) = f^\prime(0)^2$ or $f^\prime(0) = 1$. Hence for each $x$, either $f^\prime(f(x)) = 0$ or $f^\prime(x) = 1$. If $f^\prime(f(x)) = 0$ then $f^\prime(x) = 0$, so for each $x$, either $f^\prime(x) = 0$ or $f^\prime(x) = 1$.
By the Mean Value Theorem, if $f(x) \neq 0$ and $f(x) \neq x$ for some $x$, then there is some $c$ such that $f^\prime(c) = \frac{f(x)-f(0)}{x-0} = \frac{f(x)}{x} \not\in\{0,1\}$, which is impossible. So $f(x)$ is either $0$ or $x$ for every $x$. By the Intermediate Value Theorem, if $f(x) = 0$ and $f(y) = y$ (with $x<y$ and $x,y\neq 0$), then there is some $a\in (x,y)$ such that $f(a) = x$, and hence $f(a) \not\in\{0,a\}$. (The same argument works for $y>x$.) It follows that there are no such $x,y$. Therefore, either $f(x) = 0$ for all $x$, or $f(x) = x$ for all $x$.
