Solution
= Solution
For $x\in[0,1]$, the nonnegative function $x(1-x^2)$ has derivative $1-3x^2$, so its maximum occurs at $x=1/\sqrt3$ and equals
$$
\frac1{\sqrt3}\left(1-\frac13\right)=\frac{2}{3\sqrt3}.
$$
The map preserves the interval exactly when this maximum is at most one:
$$
\lambda\frac{2}{3\sqrt3}\leq1.
$$
Thus the upper limit is
$$
\boxed{\lambda_{\max}=\frac{3\sqrt3}{2}}.
$$