Ah there should be a fair number of them; for example consider
"https://en.wikipedia.org/wiki/Cantor_function"
Cantor's Function is not absolutely continuous and not lips-chitz continuous,
$$[0,1] \to [0,1]$$
But it is uniformly continuous and an even 'Holder Continuous' which is stronger again on that domain and range
$[0,1] \to [0,1]$
also see https://en.wikipedia.org/wiki/Minkowski%27s_question_mark_function
Minkowskis ? question function F, is even more peculiar: it is even 'odd', 'symmetric', with some other linear like properties, associated with lipschitz functions, and strictly increasing, continuous.
$$[0,1] \to [0,1]$$
This function agrees $F(x)=x$, a linear (lip-schitz exponent 1) function on at least five places (five points where F(x)=x); five fixed points,
$$F(1)=1\,,F(0.5)=0.5\,\, ,F(0)=0$$
and on at least two other elements of the domain.
I believe that 'F' biject-ive, and may possess a stronger form of holder continuity again. But F is not absolutely continuous on that interval and thus not Lip-schitz continuous
Following the definition of Lipschitz condition for $f$, we need to show that for some constant $K>0 $ one has
$$
|f(x) - f(y)| \leq K |x- y| \text{ for all } x,y \in [-1,1].
$$
Now take $y = 1$, then $f(y) = 0$ and the above becomes
$$
\tag{1} \sqrt{1 - x^2} \leq K |x - 1|, \text{ for all } x \in [-1,1].
$$
However,
$$
\lim\limits_{x\to 1-}\frac{\sqrt{1 - x^2}}{1-x} = \lim\limits_{x\to 1-}\frac{\sqrt{1 + x}}{\sqrt{1 - x}} = + \infty,
$$
hence no $K$ satisfies $(1)$.
Best Answer
Consider the function $$f(t):=t^2\qquad(-\infty<t<\infty)\ .$$ This function is not Lipschitz continuous on its domain, but $$\nabla f(t)=f'(t)=2t$$ is.