The ordinary proof that 0 derivative implies constant rests on the mean value theorem. Your function $g(x)$ is the "symmetric derivative" of $f$, so we should look for a "quasi-mean value theorem" for symmetric derivatives. A quick google search came up with the paper
http://www1.au.edu.tw/ox_view/edu/tojms/j_paper/Full_text/Vol-27/No-3/27%283%298-5%28279-301%29.pdf
Here it is shown (Theorem 2) that if $f$ is continuous on $[a,b]$ and symmetric differentiable on $(a,b)$ with symmetric derivative $f^s$, then there are points $\xi,\eta\in (a,b)$ with
$$f^s(\eta) \leq \frac{f(b)-f(a)}{b-a} \leq f^s(\xi).$$
Clearly then if $f^s = 0$ we must have that $f$ is constant.