16 Let the function $f(x)=x(1+x)^{2}, x \in(-\infty, 0]$.
(1) Find the extremum points of $f(x)$;
(2) For any $a<0$, let $F(a)$ denote the point where $f(x)$ attains its minimum value on $[a, 0]$, find the minimum value of $k=$ $\frac{F(a)}{a}$.
16
(1)
$$
\begin{aligned}
f^{\prime}(x) & =(1+x)^{2}+2 x(1+x) \\
& =(1+x)(1+3 x) \\
& =0
\end{aligned}
$$

Solving, we get $x=-1$, $x=-\frac{1}{3}$. When $x<-\frac{1}{3}$, $f^{\prime}(x)>0$; when $-1<x<-\frac{1}{3}$, $f^{\prime}(x)<0$; when $x>-\frac{1}{3}$, $f^{\prime}(x)>0$.

When $-\frac{4}{3} \leqslant a \leqslant-\frac{1}{3}$, $F(a)=-\frac{4}{27}$, then
$$
k=\frac{-\frac{4}{27}}{a} \geqslant \frac{-\frac{4}{27}}{-\frac{4}{3}}=\frac{1}{9},
$$

In particular, when $a=-\frac{4}{3}$, we have $k=\frac{1}{9}$; when $-\frac{1}{3}<a<0$, $F(a)=a^3+2a^2+a$, and $F(a)$ is increasing, so $k=\frac{F(a)}{a}>F'(-\frac{1}{3})=\frac{1}{9}$.

Therefore, for any $a<0$, the minimum value of $k$ is $\frac{1}{9}$, at which $a=-\frac{4}{3}$.