Let us consider the identity function $$f:(\mathbb{R},d)\to (\mathbb{R},d_{usual}) $$ $$f:x \to x$$ Here we are considering $d(x,y)=|(x)^3-(y)^3|$
Is the function $f$ uniformly continuous on closed and bounded interval?
I am looking for an example of function $f$ which is not uniformly continuous on a closed and bounded interval but it is continuous.
In this answer I assume that $\tan^{-1}$ denotes the arctangent $\arctan$.
No, your example does not work, because $\arctan$ is Lipschitz continuous. Indeed for all $x, y \in \Bbb R$ $$|\arctan x - \arctan y| \le |x-y|$$
this implies that the identity map $f: \Bbb R \to ( \Bbb R , d)$ is Lipschitz (hence uniformly continuous).
Anyway, if you want to build an example of a continuous function which is not uniformly continous on a closed and bounded interval, you are trying to find a counterexample to Heine-Cantor theorem. In particular you need to find a distance defined on the real line with the property that closed and bounded intervals are not compact.