I wanted to post here my solution and check if I'm right.
Also, I wanted to know how can you prove it with "Formal" way with Epsilon-Delta (I saw some posts but most people jump immediately to a solution).
My try :
I will take the sequences $x_n=n$ and $y_n=n+\frac {1}{n}$
$|x_n-y_n|=|n-n-\frac {1}{n}|=\frac {1}{n} \rightarrow 0$ (when $n \rightarrow \infty$)
but $|f(x_n)-f(y_n)|=|n^3-(n+\frac{1}{n})^3|=\cdots=3n+\frac {3}{n} + \frac {1}{n^3} \rightarrow \infty$ (when $n\rightarrow \infty$).
which shows that the function is not uniformly continuous over $\mathbb{R}$.
Yes, that's fine.
If you want's a $\varepsilon-\delta$ approach, it's almost the same thing. You want to prove that $f$ is not uniformly continuous, which means that$$(\exists\varepsilon>0)(\forall\delta>0)(\exists x,y\in\Bbb R):|x-y|<\delta\wedge\bigl|f(x)-f(y)\bigr|\geqslant1.$$Take $\varepsilon=1$. Now, if $\delta>0$, you take $n\in\Bbb N$ such that $\frac1n<\delta$. Now you take $x=n$ and $y=n+\frac1n$ and your computations show that, indeed,$$\bigl|f(x)-f(y)\bigr|\geqslant1.$$