If the difference quotient $\frac{f(y)-f(x)}{y-x}$ has a limit along a line $(x,y)\to(c,c)$, does the ordinary derivative $f'(c)$ exist?

The following is a well-known fact:

Claim. Let $f$ be a real-valued function defined on a neighboorhood of $0$. Suppose that $f$ is continuous at $0$ and there exist $r$ and $L$ with $|r| \neq 1$ such that $$\lim_{x\to0} \frac{f(rx) - f(x)}{(r-1)x} = L.$$ Then $f$ is differentiable at $0$ and $f'(0)=L$.

Proof. By replacing $r$ by $1/r$ if necessary, we may assume that $0 < |r| < 1$. Define

$$ \varphi(x) = \begin{cases} \dfrac{f(x) - f(rx)}{(1-k)x}, & \text{if $x \neq 0$}, \\ L, & \text{if $x = 0$}. \end{cases} $$

Then $\varphi$ is continuous at $0$. Pick a neighborhood $U$ of $0$ and $M > 0$ such that $U$ lies in the domain of $f$ and $\left| \varphi(x) \right| \leq M$ on $U$. Then

$$ \frac{f(x) - f(r^n x)}{x} = \sum_{k=1}^{n} (1-r)r^{k-1} \varphi(r^{k-1}x). $$

Since each term is bounded by $M(1-r)|r|^{k-1}$ on $I$ and $\sum_{k=1}^{\infty} M(1-r)|r|^{k-1} < \infty$, the right-hand side converges uniformly as $n\to\infty$ on $U$ by the Weierstrass M-test. So if $x \in U$, then by letting $n\to\infty$, we get

\begin{align*} \frac{f(x) - f(0)}{x} &= \lim_{n\to\infty} \frac{f(x) - f(r^n x)}{x} \\ &= \lim_{n\to\infty} \sum_{k=1}^{n} (1-r)r^{k-1} \varphi(r^{k-1}x) \\ &= \sum_{k=1}^{\infty} (1-r)r^{k-1} \varphi(r^{k-1}x). \end{align*}

Now we take limit as $x \to 0$. By the uniform convergence and the existence of term-wise limit, we can interchange the order of summation and limit, obtaining

\begin{align*} \lim_{x \to 0} \frac{f(x) - f(0)}{x} &= \sum_{k=1}^{\infty} \lim_{x \to 0} (1-r)r^{k-1} \varphi(r^{k-1}x) \\ &= \sum_{k=1}^{\infty} (1-r)r^{k-1} L \\ &= L. \end{align*}

This completes the proof. $\square$

Remark. The notion of uniform convergence is not a necessity here. Rather, it is utilized in order to make the proof more transparent.