$f$ 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?

Note that for all $h$ we have
$$4h^{2} f''(x)=2hf'(x+h)-2hf'(x-h)=f(x+2h)-2f(x)+f(x-2h).$$
Deriving again with respect to $x$ gives
$$8h^{3} f'''(x)=2hf'(x+h)-4hf'(x)+2hf(x-2h)=f(x+3h)-3f(x+h)+3f(x-h)-f(x-3h)$$
Note that $f(x+3h)-f(x-3h)=6hf'(x)$ and $-3f(x+h)+3f(x-h)=-6hf'(x)$, hence 
$$8h^{3} f'''(x)=0.$$
As this holds for all $h$ we find that $f'''(x)=0$ and thus $f(x)=ax^{2}+bx+c$. 
