$f$ has the form $f(x) =ax^2+bx+c$. Every differentiable function $f:R \rightarrow R$ with the property that $(2h)f′(x) =f(x+h)−f(x−h)$ for all $x \in R$ and all $h$ has the form $f(x) =ax^2+bx+c$.
I would like to get some help on this one. Taking partial derivatives was an idea, but is that legal?

We have $f'(x)={1\over2}\bigl(f(x+1)-f(x-1)\bigr)$ for all $x$. Since here
 the RHS  is differentiable it follows that $f'$ is differentiable as well; in fact $f\in C^\infty$. We now differentiate
$2 h f'(x)=f(x+h)-f(x-h)$ two times with respect to $h$ and obtain
$$0=f''(x+h)-f''(x-h)\qquad\forall\,x\quad\forall\, h\ .$$
This implies $f''(u)=f''(v)$ for all $u$ and $v$, i.e., $f''(x)$ is a constant, say $2a$. This in turn implies $f'(x)=2ax+b$ and $f(x)=ax^2+bx+c$ for constant $a$, $b$, $c$.
