show that $f(x) = \frac{x^2}{ 1 + \vert x \vert} $ is uniformly continuous on $\mathbb{R}$ I started proving it by saying that on [-1;1] the map is continuous and so we have uniform continuity.
Then, I want to show that f is uniform continuous on [1; $\infty$[ and ]-$\infty$ ; -1].
I wanted to do it with the definition, I have the following expression: $\forall x,y \in [1; \infty$[ \
$\mid f(x) - f(y) \mid = \mid$ $\frac{x^2}{1 + \mid x \mid} - \frac{y^2}{ 1 + \mid y \mid} \mid = 
\mid \frac{x^2 - y^2 + x^2y -y^2x}{(1+x)(1+y)} \mid = \mid \frac{(x - y)( x+ y + xy)}{(1+x)(1+y)} \mid$
I don't know how to conclude. I also have to show that the result stays true for $x \in [1; \infty$[ and
y $\in  ]-\infty ; -1]$ ?

You are almost there as $f$ is even and for $x,y \ge 0$
$$\begin{aligned}\left\vert f(x) - f(y) \right\vert &= \left\vert \frac{x^2}{1 + \vert x \vert} - \frac{y^2}{ 1 + \vert y \vert} \right\vert\\
&= 
\left\vert \frac{x^2 - y^2 + x^2y -y^2x}{(1+x)(1+y)} \right\vert\\
&= \left\vert \frac{(x - y)( x+ y + xy)}{(1+x)(1+y)}\right\vert\\
&\le \vert x - y \vert \end{aligned}$$
knowing that $0 \le x+ y + xy \le (1+x)(1+y)$
