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? Given a function $f:\mathbb R\to\mathbb R$, we define the difference quotient function
$$q(x,y)=\frac{f(y)-f(x)}{y-x}$$
for all $(x,y)\in\mathbb R^2$ not on the diagonal line $x=y$.
The ordinary derivative $f'(c)$ is defined as a limit of $q$ along a horizontal ($y=c$) or vertical ($x=c$) line through $(c,c)$.
The symmetric derivative is a limit along a diagonal line $y-c=c-x$.
The left derivative is a limit along a horizontal ray $y=c,\,x<c$.
The right derivative is a limit along a vertical ray $x=c,\,y>c$.
The strong derivative is the limit of $q$ at $(c,c)$, not along any particular path.

If the ordinary derivative exists, then $q(x,y)\to f'(c)$ along any line through $(c,c)$, or in any region (a "cone") separated from the diagonal by lines through $(c,c)$:
$$q(x,y)=\frac{y-c}{y-x}\cdot\frac{f(y)-f(c)}{y-c}+\frac{c-x}{y-x}\cdot\frac{f(c)-f(x)}{c-x}$$
$$=\frac{y-c}{y-x}\cdot q(c,y)+\frac{c-x}{y-x}\cdot q(x,c).$$
Note that the coefficients sum to $1$, and we're given $q(c,x)-f'(c)\to0$ as $x\to c$, so
$$q(x,y)-f'(c)=\frac{y-c}{y-x}\big(q(c,y)-f'(c)\big)+\frac{c-x}{y-x}\big(q(x,c)-f'(c)\big)$$
$$\to0,$$
provided that the coefficients $\frac{y-c}{y-x}$ and $\frac{c-x}{y-x}$ are bounded.

If $q$ has a limit $f^*(c)$ along a line, say $y-c=k(x-c)$ with $0\neq|k|\neq1$, does the derivative exist?
To simplify the notation, let's assume $c=f(c)=f^*(c)=0$. We're given, as $x\to0$,
$$\frac{f(kx)-f(x)}{kx-x}=\frac{1}{k-1}\cdot\frac{f(kx)-f(x)}{x}\to0,$$
and we want to know whether $\frac{f(x)}{x}\to0$.
There are discontinuous counter-examples: Let $f(k^n)=1$ for $n\in\mathbb Z$, and otherwise $f(x)=0$; then $q(x,kx)=0\to0$, but $q(0,x)\not\to0$. So let's assume that $f$ is continuous at $c$, and maybe in a neighbourhood of $c$.

Since $q(x,y)=q(y,x)$ is symmetric, the limit along a line with slope $k$ is the same as with slope $1/k$. So, without loss of generality, $0<|k|<1$.
If the limit is $0$ for two lines with slopes $k$ and $l$, then it's also $0$ for a line with slope $k\cdot l$:
$$\lim_{x\to0}\frac{f(klx)-f(x)}{x}=\lim_{x\to0}\frac{f(klx)-f(lx)+f(lx)-f(x)}{x}$$
$$=l\cdot\lim_{lx\to0}\frac{f(klx)-f(lx)}{lx}+\lim_{x\to0}\frac{f(lx)-f(x)}{x}=0.$$
Thus, the limit is $0$ for any line with slope $k^n$ where $n\in\mathbb N$.
Now the derivative is
$$f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x}$$
$$=\lim_{x\to0}\frac{f(x)-f(\lim_{n\to\infty} k^nx)}{x}$$
and we assumed that $f$ is continuous at $0$:
$$=\lim_{x\to0}\frac{f(x)-\lim_{n\to\infty}f(k^nx)}{x}$$
$$=\lim_{x\to0}\lim_{n\to\infty}\frac{f(x)-f(k^nx)}{x}$$
$$\overset?=\lim_{n\to\infty}\lim_{x\to0}\frac{f(x)-f(k^nx)}{x}$$
$$=\lim_{n\to\infty}(0)=0.$$
Is swapping the limits valid here?
 A: 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.
A: Thanks to @ParamanandSingh for the essence of this answer: https://math.stackexchange.com/a/1864092
From the definition of the limit, for any $\varepsilon>0$, there is $\delta>0$ such that, for all $0<|x|<\delta$,
$$\left|\frac{f(kx)-f(x)}{x}\right|<\varepsilon.$$
Since $|k|<1$ and thus $|k^nx|<|x|<\delta$ for $n\in\mathbb N$, this is also true if we replace $x$ with $k^nx$:
$$\left|\frac{f(k^{n+1}x)-f(k^nx)}{x}\right|<|k|^n\varepsilon.$$
Now we make a telescoping series:
$$\left|\frac{f(k^mx)-f(x)}{x}\right|=\left|\frac{\sum_{n=0}^{m-1}\big(f(k^{n+1}x)-f(k^nx)\big)}{x}\right|$$
$$\leq\sum_{n=0}^{m-1}\left|\frac{f(k^{n+1}x)-f(k^nx)}{x}\right|$$
$$<\sum_{n=0}^{m-1}|k|^n\varepsilon$$
$$=\frac{1-|k|^m}{1-|k|}\,\varepsilon$$
and let $m\to\infty$ (using continuity of $f$ at $0$) to get
$$\left|\frac{f(0)-f(x)}{x}\right|<\frac{\varepsilon}{1-|k|}$$
for all $0<|x|<\delta$. Thus, given any $\varepsilon'>0$, we can define $\varepsilon=(1-|k|)\varepsilon'$ to get a neighbourhood of $0$ within which $\left|\frac{f(x)-f(0)}{x}\right|<\varepsilon'$. That is, $f'(0)=0$.
