Find example: $\lim_{h\to0}\frac{f(x_0+\alpha h)-f(x_0-\beta h)}{(\alpha+\beta)h}\ \ \text{exists }\ \ \not\!\!\!\implies\ f'(x_0)\ \ \text{exists}$ 
Assume that $f:[a,b]\to\mathbb R$ is a continuous function. For $x_0\in (a,b)$ and $\alpha,\beta>0$, we define the (asymetric) difference quotient
$$(\Delta_{\alpha,\beta,h}f)(x_0):=\frac{f(x_0+\alpha h)-f(x_0-\beta h)}{(\alpha+\beta)h},\qquad h\neq 0.$$
Given $\alpha,\beta>0$ with $\alpha\neq \beta$, I want to find an example of $f$ such that
$$\lim_{h\to0}(\Delta_{\alpha,\beta,h}f)(x_0)\ \ \text{exists }\ \  \not\!\!\!\implies\ f'(x_0)\ \ \ \text{exists}.$$

This is a problem that I came up with while I'm preparing my recitation class as a teaching assistant in Analysis (I) course.
If $f'(x_0)$ exists, we can prove that $\lim_{h\to0}(\Delta_{\alpha,\beta,h}f)(x_0)=f'(x_0)$ by Taylor's theorem: It follows from $f(x_0+\alpha h)=f(x_0)+\alpha f'(x_0)h+o(h)$ and $f(x_0-\beta h)=f(x_0)-\beta f'(x_0)h+o(h)$ that $(\Delta_{\alpha,\beta,h}f)(x_0)=f'(x_0)+o(1)$ as $h\to0$, hence $\lim_{h\to0}(\Delta_{\alpha,\beta,h}f)(x_0)=f'(x_0)$.
Now, a natural question appears: What about the reverse problem? If $\alpha=\beta$, then for $f(x)=|x|$ and $x_0=0$ we have $(\Delta_{\alpha,\beta,h}f)(0)=0$ for all $h\neq0$, but $f'(0)$ doesn't exist.
However, for $\alpha\neq \beta$, I failed to find an example. I found that if $\alpha\neq \beta$, and if $f_-'(x_0)$, $f_+'(x_0)$ both exist, then we must have $$\lim_{h\to0}(\Delta_{\alpha,\beta,h}f)(x_0)\ \ \text{exists }\ \  \implies\ f'(x_0)\ \ \ \text{exists}.$$
So, for any example in which $\alpha\neq \beta$ and $\lim_{h\to0}(\Delta_{\alpha,\beta,h}f)(x_0)\ \ \text{exists }\ \  \not\!\!\!\implies\ f'(x_0)\ \ \ \text{exists}$, there must be one of $f_-'(x_0)$ and $f_+'(x_0)$ that doesn't exist. (Note that $f$ should be continuous in this post.)
Any help would be appreciated!
 A: I have to apologize... Thinking for more time leads to the conclusion that we cannot find any desired example. Let me explain.

Proposition. Let $f$ be a function defined near $x_0\in\mathbb R$ and $\alpha, \beta\in\mathbb R$  be such that $|\alpha|\neq |\beta|$. Assume that $f$ is continuous at $x_0$, and
$$\lim_{h\to0}\frac{f(x_0+\alpha h)-f(x_0+\beta h)}{(\alpha-\beta)h}=A\in\mathbb R.$$
Then $f$ is differentiable at $x_0$ and $f'(x_0)=A$.

Proof. Without loss of generality, we assume that $x_0=0$, $|\alpha|>|\beta|$ and $A=0$. Let $k=\frac\beta\alpha$, then $|k|<1$ and we have
$$\lim_{h\to0}\frac{f(h)-f(kh)}{(1-k)h}=0.$$
We are going to prove that $f'(0)=0$. For any $\epsilon>0$, there exists $\delta>0$ such that $$\left|\frac{f(h)-f(kh)}{(1-k)h}\right|<\epsilon,\qquad 0<|h|<\delta.$$
Now, for $0<|h|<\delta$, we have $0<|k^nh|<\delta$ for all $n\in\mathbb N_{>0}$, hence
$$\left|\frac{f\left(k^{n-1}h\right)-f\left(k^{n}h\right)}{h}\right|<(1-k)|k|^{n-1}\epsilon,\qquad \forall n\in\mathbb N_{>0},$$
and thus
\begin{align*}
\left|\frac{f\left(h\right)-f\left(k^{n}h\right)}{h}\right|&\leq\sum_{j=1}^n\left|\frac{f\left(k^{j-1}h\right)-f\left(k^{j}h\right)}{h}\right|\\
&<\sum_{j=1}^n(1-k)|k|^{j-1}\epsilon\\
&<\frac{1-k}{1-|k|}\epsilon, \qquad \forall n\in\mathbb N_{>0}.
\end{align*}
Letting $n\to\infty$, the continuity of $f$ at $x=0$ gives that
$$\left|\frac{f\left(h\right)-f\left(0\right)}{h}\right|\leq\frac{1-k}{1-|k|}\epsilon,\qquad 0<|h|<\delta.$$
Therefore, $f$ is differentiable at $x=0$ and $f'(0)=0$. The proof is complete.
A: I think this is too long for a comment, but I'm not sure how complete this is. Hopefully it is helpful.
Consider
\begin{align}
\lim_{h \to 0^{\pm}} \frac{f(x_0+\alpha h)-f(x_0-\beta h)}{(\alpha+\beta)h} & = \lim_{h \to 0^{\pm}} \frac{f(x_0+\alpha h) - f(x_0) + f(x_0)-f(x_0-\beta h)}{(\alpha+\beta)h} \\ & = \frac{\alpha}{\alpha + \beta} f'(x_0^{\pm}) + \frac{\beta}{\alpha + \beta} f'(x_0^{\mp}).
\end{align}
Now since $\lim_h\to0 \Delta_{\alpha,\beta,h}$ exists, we must have equality when taking the limit from either direction. This leads to
$$ (\alpha-\beta) f'(x_0^+) = (\alpha - \beta)f'(x_0^-).$$
By hypothesis if $\alpha \neq \beta$, we may cancel and have shown that $f'(x_0)$ exists.
