Local Lipschitz constants and a slightly weaker concept Let $(X, d)$ be a metric space and $f:X \rightarrow \mathbb R$ be a function such that
$$\lim_{r \rightarrow 0} \sup_{0 < y < B(x; r)} \frac{|f(x) - f(y)|}{d(x, y)}$$
exists for some $x \in X$.
Does
$$\lim_{r \rightarrow 0} \sup_{0 < y < B(x; r)} \frac{|f(x) - f(y)|}{r}  $$
exists in this case?
Since $d(x, y) < r$, it is clear that
$$\lim_{r \rightarrow 0} \sup_{0 < y < B(x; r)} \frac{|f(x) - f(y)|}{d(x, y)} \leq \lim_{r \rightarrow 0} \sup_{0 < y < B(x; r)} \frac{|f(x) - f(y)|}{r} .$$
Then when does the equal sign hold?
 A: To observations:

*

*The notation $0 < y < B(x; r)$ is unusual.

*Although the first limit exists always (maybe equals $\infty$) as a limit of monotone funcion, the second one not always. We therefore ought to consider upper and lower limits.

*There's a mistake in the question. The condition $d(x,y)<r$ implies that
$$\lim_{r \rightarrow 0} \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{d(x, y)} \color{red}{\geq} \limsup_{r \rightarrow 0} \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{r} .$$

My example is based on noncontinuous function. If a continuous function is needed one can change a bit the definition of $f$ to be partially linear.
Let $f\colon\Bbb R\to\Bbb R$, $f(2^{-n})=2^{-n}$ and $f(x)=0$ for other points.
Then $\sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{d(x,y)}=1$ for any $r>0$ and so the first limit is equal to $1$. On the other hand, for $r=2^{-n}$
$$
\sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{r} = 
\sup_{y\in B(x; r)\setminus\{x\}} \frac{f(y)}{r} = 
\sup_{y\in B(x; r)\setminus\{x\}} \frac{2^{-n-1}}{2^{-n}} =\frac 12. 
$$
Therefore $$ \liminf_{r \rightarrow 0} \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{r}\leq \frac 12,$$
which shows that the second limit isn't equal to $1$.

We can prove that the second limit doesn't exists in our case. Namely, for $r=2^{-n}+4^{-n}$ we have $2^{-n}<r<2^{-n+1}$ and therefore
$$
\sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{r} = 
\sup_{y\in B(x; r)\setminus\{x\}} \frac{f(y)}{r} = 
\sup_{y\in B(x; r)\setminus\{x\}} \frac{2^{-n}}{2^{-n}+4^{-n}} \to 1 \ \ (n\to\infty),
$$
so $$\limsup_{r \rightarrow 0} \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{r}\geq 1.$$
Final remark. It can be shown that
$$\lim_{r \rightarrow 0} \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{d(x, y)} \color{red} = \limsup_{r \rightarrow 0} \sup_{y\in B(x; r)\setminus\{x\}} \frac{|f(x) - f(y)|}{r} .$$
