The functional equation $f(x)+f(y)+f(z)+f(x+y+z)=f(x+y)+f(y+z)+f(z+x)$
We first use divided differences to show that $f$ is three times differentiable. In fact, continuity is not needed. Let $$ \delta_h f(x)=f(x+h)-f(x). $$ Then writing the functional equation with $x$ replaced by $x+h$ and subtracting the original equation from it, we derive $$ \delta_hf(x)+\delta_hf(x+y+z)=\delta_hf(x+y)+\delta_hf(z+x). $$ Now doing the same with $y$, and subsequently with $z$, we get $$ \delta_\ell\delta_k\delta_hf(x)=0, $$ for all $h,k,\ell\in\mathbb{R}$ and $x\in\mathbb{R}$. In particular, the following limit exists $$ \delta_k\delta_hf'(x)=\lim_{\ell\to0}\frac{\delta_\ell\delta_k\delta_hf(x)}{\ell}=0. $$ Proceeding similarly, we conclude that $f'''(x)$ exists and equal to $0$ everywhere.
This means that $f$ is of the form $$ f(x)=ax^2+bx+c. $$ We then simply check if each of $1,x$, and $x^2$ satisfies the functional equation, and conclude that the general solution is $$ f(x)=ax^2+bx. $$
All continuous solutions are of the form $f(x) = a x + b x^2$.
For given $z$, let $g(x) = f(x+z) - f(x) - f(z)$. The functional equation becomes $g(x+y) = g(x) + g(y)$. This is the Cauchy functional equation, and it is known that its only continuous solutions are $g(x) = c x$. Of course $c$ can depend on $z$. Now we need to solve $f(x+z) - f(x) - f(z) = c(z) x$. Since the left side is symmetric in $x$ and $z$, $c(z) x = c(x) z$, so $c(z) = k z$ for some constant $k$. Taking $z=-x$, and using $f(0)=0$, we get $- f(x) - f(-x) = - k x^2$, or $f(x) + f(-x) = k x^2$.
Note that if $f(x)$ is a solution of our equation, so is $f(-x)$, and by linearity so are the even and odd parts $ (f(x) + f(-x))/2$ and $(f(x) - f(-x))/2$. Thus it suffices to consider the two cases $f$ even and $f$ odd.
If $f$ is even, $f(x) + f(-x) = 2 f(x) = k x^2$, so $f(x) = (k/2) x^2$.
If $f$ is odd, $f(x) + f(-x) = 0$ so $k=0$. now we have $f(x+z) - f(x) - f(z) = 0$, which is again Cauchy's functional equation, and so $f(x) = a x$ for some constant $a$.
Consider the left and right as functions of three variables.
- Take $\partial / \partial x$ of both sides (keeping track of the chain rule when necessary):
$$ f'(x) + f'(x+y+z) = f'(x+y) + f'(z+x) $$
- Take $\partial / \partial y$ of both sides of the result:
$$ f''(x+y+z) = f''(x + y) $$
- Take $\partial / \partial z$ of both sides of that:
$$ f'''(x+y+z) = 0 $$
Now substitute $y=z=0$ to find that $f'''(x)=0$. This shows that the only possible functions would be of the form $f(x) = ax^2 + bx + c$ for constants $a, b, c$. Then apply Robert's observation (see comments above) that $f(0) = 0$ (hence $c=0$), and that the set of functions form a linear space (hence all choices of $a, b \in \mathbb{R}$ are valid).
Hope this helps!
Let $f: \mathbb R \to \mathbb R$ be a continous solution of the functional equation and let $g(x) := ax^2+bx$ where $a$ and $b$ are chosen in such a way that $f(-1) = g(-1)$ and $f(1) = g(1)$. Then $\tilde f := f-g$ is also a solution and we have $\tilde f(-1) = \tilde f(1) = 0$. We will show $\tilde f \equiv 0$, so that $f = g$.
By replacing $f$ with $\tilde f$ way may assume that $f$ is a solution with $f(-1) = f(1) = 0$, and we will show $f \equiv 0$. Let $Z = \{x \in \mathbb R: f(x) = 0 \}$. A priori we have $\{-1,+1\} \subseteq Z$. Plugging $(1,-1,x)$ into the functional equation yields $$f(x) = \frac{f(x+1) + f(x-1)}{2},$$ which implies $\mathbb Z \subseteq Z$. If we can show $$2x,x \in Z \Rightarrow \frac{x}{2} \in Z$$ we are finished since then every dyadic rational $a/2^n$ is in $Z$ and these are dense in $\mathbb R$ (here the continuity of $f$ is needed). So let $2x \in Z$ and $x \in Z$. Plug $(\frac{x}{2},\frac{x}{2},x)$ into the functional equation to get $$2f\left(\frac{x}{2}\right) + f(x) + f(2x) = f(x) + 2f\left(\frac{3}{2}x\right),$$ hence $f\left(\frac{x}{2}\right) = f\left(\frac{3}{2}x\right)$. Then plugging in $(\frac{x}{2},\frac{x}{2},\frac{x}{2})$ yields $$3f\left(\frac{x}{2}\right) + f\left(\frac{3}{2}x\right) = 3f(x),$$ hence $f(\frac{x}{2}) = 0$ and $\frac{x}{2} \in Z$.