Hint. Let $f(x)=3x^3/(1+x^2)$ then
$$f(x)-f(y)=3\cdot \frac{x^2y^2+x^2+xy+y^2}{(x^2+1)(y^2+1)}\cdot (x-y).$$
Show that
$$0\leq \frac{x^2y^2+x^2+xy+y^2}{(x^2+1)(y^2+1)}\leq 2,$$
that is
$$0\leq x^2y^2+x^2+xy+y^2\leq 2x^2y^2+2x^2+2y^2+2.$$
Hence $|f(a_n)-f(a_m)|\leq 6 |a_n-a_m|$.
P.S. More generally if $(a_n)_n$ is a Cauchy sequence then it is bounded: there is $M>0$ such that $|a_n|\leq M$ for all $n\in\mathbb{N}$. Moreover
if $f$ is a $C^1(\mathbb{R})$ function then $f'$ is bounded over the compact set $[-M,M]$. Hence, by the Mean Value Theorem, if $a_n\not= a_m$, then there exists $t\in [-M,M]$ (which depends on $a_n$ and $a_m$) such that $f(a_n)-f(a_m)=f'(t)(a_n-a_m).$
Hence
$$|f(a_n)-f(a_m)|\leq C |a_n-a_m|$$
where $C:=\max_{x\in [-M,M]}|f'(x)|$.