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?