 Solution

For the iteration of a map $f(x)=\lambda x(1-x^2)$, a fixed point satisfies $x[1-\lambda +\lambda x^2]=0$. If $0<\lambda <1$, the bracket is strictly positive, so the sole fixed point is $0$. Its multiplier of a periodic orbit of an iteration is $f^{\prime }(0)=\lambda$, whose modulus is below one. Thus $\boxed{x=0\text{ is locally asymptotically stable}}$ by fixed point stability for an iteration.