Anti-symmetric derivative implies differentiability Wikipedia defines Symmetric derivative as a limit $$\lim_{h\to 0} \frac{f(x+h)-f(x-h)} {2h}$$
which doesn't imply differentiability. An obvious example is $f(x)=|x|$ or $f=\chi_{\mathbb Q}$ 
I think a slightly different, unbalanced form of the limit guarantees differentiability. 
Especially the following limit
$$\lim_{h\to 0} \frac{f(a+2h)-f(a-h)} {3h}=\gamma$$
and the continuity of $f$ at $a$ implies $f'(a)=\gamma$.   
My question is,   

  
*
  
*Is the following proof valid?   
  
*Is it possible to prove an analogous extension with the condition $$\lim_{h\to 0} \frac{f(x+ph)-f(x-qh)} {(p+q)h} = \gamma$$
  while $p,q$ are different positive real values?   
  

(Proof)   
Define 
$$G(h)=\begin{cases} 
{\frac{f(a+2h)-f(a-h)} {3h}-\gamma}\quad h\neq 0 \\
0 \quad\quad\quad \quad \quad\quad\quad\quad h=0\\
\end{cases}
$$ 
and substitute $h$ with $\frac 12 (-\frac 12)^kh = x_k h$.
$$G(x_kh)=\frac{f(a+ (-\frac 12)^kh)-f(a+(-\frac 12)^{k+1}h)} {3x_k h}-\gamma$$ 
By multiplying $3x_k$ both side and adding up from $k=0$ to $n-1$,  
$$3\sum_{k=0}^{n-1}{x_k G(x_k h)}=\sum_{k=0}^{n-1}(\frac{f(a+ (-\frac 12)^kh)-f(a+(-\frac 12)^{k+1}h)} h)- (1-(-\frac 12)^n)\gamma
$$
RHS become $$\frac{f(a+h)-f(a-(1/2)^n)} h - (1-(-\frac 12)^n)\gamma\quad \to \quad\frac{f(a+h)-f(a)} h -\gamma$$
by continuity.   
So we have to show that the left side is zero. Since $|G(x)|<M$ for some $0$ neighborhood, Weierstrass M-test guarantees that $\sum_k ^\infty{x_k G(x_kh)}$ is continuous at $h=0$ and obviously its value is zero.
 A: I think that your proof is correct, and also works for the generalized assumption. Since $(p, q)$ can be scaled by a common factor, it can be stated as follows:

Let $I \subset \Bbb R$ be an interval and $f: I \to \Bbb R$ a function which is continuous at $a \in I$. If
  $$\lim_{h \to 0} \frac{f(a+h)-f(a+ch)}{(1-c)h} = \gamma$$
  for some $c \in (-1, 1)$ then $f$ is differentiable at $a$, and $f'(a) = \gamma$.

For the proof I would proceed as follows: To simplify the notation one can assume that $a=0$, $f(0) = 0$, and $\gamma = 0$ so that
$$\tag{*}\lim_{h \to 0} \frac{f(h)-f(ch)}{(1-c)h} = 0 \, .$$
Then for all $n > 0$
$$
\frac{f(x)}{x} = (1-c)\sum_{k=0}^{n-1} c^k \frac{f(c^kx)-f(c^{k+1}x)}{c^k x - c^{k+1}x} + \frac{f(c^nx)}{x}
$$
From the condition $(*)$ we can conclude that for a given $\varepsilon > 0$,
$$
\left|\frac{f(x)}{x} \right| \le (1-c)\sum_{k=0}^{n-1} c^k \varepsilon + 
\left|\frac{f(c^nx)}{x} \right| 
\le  \varepsilon + \left|\frac{f(c^nx)}{x} \right|
$$
for sufficiently small $x \ne 0$. For fixed $x$ and $n \to \infty$ it follows that
$$
\left|\frac{f(x)}{x} \right| \le  \varepsilon
$$
because of the continuity of $f$ at $x=0$. Since $\varepsilon > 0$ was arbitrary, we have proved that
$$
 f'(0) = \lim_{x \to 0}\frac{f(x)}{x}  = 0 \, .
$$
