Let $f: \Bbb R^d \to \Bbb R^d$ be defined as $$f(x) = \frac{x}{\max(\langle v, x \rangle, 1)}$$ where $v$ is a $d \times 1$ vector so that exactly one component is 1 and the rest are zero.
Is $f$ Lipschitz. I could not prove it. Hence I tried proving that it is not. If $L < 1$, I can find $x,y$ such that $\|f(x) - f(y)\| \geq L\|x -y\|$. But, no success for $L > 1$.
Your function is not Lipschitz continuous in general.
To see this, take $v = e_1$ (the first standard basis vector) and $y = 2n e_2$, as well as $x = 2 e_1 + 2n e_2$. Then we have $|x - y| = 2$, and $\max \{1, \langle v, x\rangle \} = 2$ and $\max\{1, \langle v, y\rangle\} = 1$. Hence, $$ |f(x) - f(y)| = |x/2 - y| = |-n e_2 + e_1| \geq n. $$ Since this holds for every $n \in \Bbb{N}$, $f$ cannot be Lipschitz continuous.