= Solution
Put $a=2/\sqrt3$. On $[-a,a]$, the extrema of $|x(1-x^2)|$ occur at $\pm1/\sqrt3$ or the endpoints. The corresponding values are $2/(3\sqrt3)$, including the endpoint value $a(a^2-1)$. Consequently
$$
|f(x)|\leq\frac{2\lambda}{3\sqrt3}<\frac2{\sqrt3}=a\qquad(\lambda<3,\ |x|\leq a).
$$
Thus $[-a,a]$ is a <trapping region> and
$$
\boxed{|x_n|<2/\sqrt3\text{ for all }n\geq1\text{ whenever }|x_0|\leq2/\sqrt3.}
$$
It contains a neighbourhood of the origin. Sign changes are possible because $f(x)$ has opposite sign to $x$ for $1<|x|<a$.
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