16. (12 points) Let the function
$$
f(x)=x(1+x)^{2}(x \in(-\infty, 0]) \text {. }
$$
(1) Find the critical 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) From
$$
\begin{array}{l}
f^{\prime}(x)=(1+x)^{2}+2 x(1+x) \\
=(1+x)(1+3 x)=0,
\end{array}
$$

we solve for $x_{1}=-1, x_{2}=-\frac{1}{3}$.
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 $a < -\frac{4}{3}$,
$$
F(a)=f(a)=a^{3}+a^{2}+a, \quad k=\frac{a^{3}+a^{2}+a}{a}=a^{2}+a+1>\frac{1}{9} \text {; }
$$

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

In particular, when $a=-\frac{4}{3}$, we have $k=\frac{1}{9}$;
When $-\frac{1}{3}< a < 0$,
$$
F(a)=f(a)=a^{3}+a^{2}+a, \quad k=\frac{a^{3}+a^{2}+a}{a}=a^{2}+a+1>\frac{1}{9} .
$$

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