Let $f:\mathbb{R}\rightarrow \mathbb{R}$ be a differentiable function at $0$.
We look for $$ \lim_{x,y\rightarrow 0}\frac{f(x)-f(y)}{x-y} $$
Do we have ($h=x-y$): $$ \frac{f(x)-f(y)}{x-y} = \frac{f(y+h)-f(y)}{h} \longrightarrow_{h,y\rightarrow 0} f'(0) $$
?
No, this is not always true. For example consider the function $f: \mathbb{R} \to \mathbb{R}$ defined as $$ f(x) = \begin{cases} x^2 \sin(\frac{1}{x}) & \text{if } x \neq 0,\\ 0 & \text{if } x = 0. \end{cases} $$ It can be verified that $f$ is differentiable on $\mathbb{R}$, and that $$ f'(x) = \begin{cases} 2x\sin(\frac{1}{x})-\cos(\frac{1}{x}) & \text{if } x \neq 0,\\ 0 & \text{if } x = 0. \end{cases} $$ The limit of interest is however not defined. If it would be defined, we could namely take the limit over the line x = 2y, and see that the limit would have to equal $$ \lim_{y \to 0} \frac{4y^2 \sin\left(\frac{1}{2y}\right) - y^2\sin\left(\frac{1}{y}\right)}{y^2} = \lim_{y \to 0} \left(4\sin\left(\frac{1}{2y}\right) - \sin\left(\frac{1}{y}\right)\right), $$ which is not defined.