Does a such condition imply differentiability? Let function $f:\mathbb{R}\to \mathbb{R}$ be such that 
$$
 \lim_{\Large{(y,z)\rightarrow (x,x) \atop y\neq z}} \frac{f(y)-f(z)}{y-z}=0.
$$
Is it then $f'(x)=0$ ?

What is $f'(x)=\displaystyle\lim_{y\to x}\dfrac{f(y)-f(x)}{y-x}$?

Let $\epsilon>0$. From the hypothesis it follows that exists a $\delta>0$ such that if $y\neq z$ and $0<\|(y,z)-(x,x)\|<\delta$ then $\left|\dfrac{f(y)-f(z)}{y-z}\right|<\epsilon$.

Now if $0<|y-x|<\delta$ then $0<\|(y,x)-(x,x)\|=|y-x|<\delta$ and $y\neq x$. 
Therefore $\left|\dfrac{f(y)-f(x)}{y-x}\right|<\epsilon$.
It follows that $f'(x)=\displaystyle\lim_{y\to x}\dfrac{f(y)-f(x)}{y-x}=0.$
