Let $A \subset \mathbb{R}^n$ and $f: A \to \mathbb{R}$.
- $f$ is stable at $c \in A$ if there exist $r_c > 0$ and $L_c > 0$ such that $$|f(x) - f(c)| < L_c\|x - c\|, \quad \text{for all}~ x \in B(c,r_c) \cap A.$$
- $f$ is stable around $c \in A$ if there exist $r_c > 0$ and $L_c > 0$ such that, for all $y \in B(c,r_c) \cap A$, there is $r_y > 0$ such that $$|f(x) - f(y)| < L_c\|x - y\|, \quad \text{for all}~ x \in B(y,r_y) \cap A.$$
- $f$ is lipschitz around $c \in A$ if there exist $r_c > 0$ and $L_c > 0$ such that $$|f(x) - f(y)| < L_c\|x - y\|, \quad \text{for all}~ x,y \in B(c,r_c) \cap A.$$
Note that, property (1) is usually called "lipschitz at". It is easy to see the implication $$(3) \to (2) \to (1).$$ I am doubt that the converse $(2) \to (3)$ is not true, but unable to find an counterexample. Here is my idea:
Constructing a piecewise function $$f(x) = \begin{cases} f_1(x), &\text{if}~ x \in A_1 \\ f_2(x), &\text{if}~ x \in A_2 \\ \end{cases}$$ such that $f$ is stable at $c \in A_1 \cap A_2$ and $f_1, f_2$ are lipschitz. Now, for all $y_1 \in B(c,r_c) \cap A_1$, one can find an $r_1 > 0$ small enough so $B(y_1,r_1) \subset A_1$ and $f(x) = f_1(x)$ for all $x \in B(y,r_1)$. The arguments are similar for $y_2 \in B(c,r_c) \cap A_2$. Hence, the stability of $f$ around $c$ is made by the lipschitz property of $f_1$ and $f_2$. However, one can find $x_1 \in A_1$, $x_2 \in A_2$ such that $$\dfrac{|f(x_1) - f(x_2)|}{\|x_1 - x_2\|}$$ is unbounded.