9. (16 points) Given the function
$$
\begin{array}{l}
f(x)=a x^{3}+b x^{2}+c x+d(a \neq 0), \\
\text { when } 0 \leqslant x \leqslant 1 \text {, }|f^{\prime}(x)| \leqslant 1 .
\end{array}
$$

Try to find the maximum value of $a$.
II. Notice that $f^{\prime}(x)=3 a x^{2}+2 b x+c$. From $\left\{\begin{array}{l}f^{\prime}(0)=c, \\ f^{\prime}\left(\frac{1}{2}\right)=\frac{3}{4} a+b+c, \\ f^{\prime}(1)=3 a+2 b+c,\end{array}\right.$
we get $3 a=2 f^{\prime}(0)+2 f^{\prime}(1)-4 f^{\prime}\left(\frac{1}{2}\right)$.
Then $3|a|=\left|2 f^{\prime}(0)+2 f^{\prime}(1)-4 f^{\prime}\left(\frac{1}{2}\right)\right|$ $\leqslant 2\left|f^{\prime}(0)\right|+2\left|f^{\prime}(1)\right|+4\left|f^{\prime}\left(\frac{1}{2}\right)\right| \leqslant 8$.
Therefore, $a \leqslant \frac{8}{3}$.
Furthermore, since $f(x)=\frac{8}{3} x^{3}-4 x^{2}+x+m$ (where $m$ is a constant) satisfies the given conditions, the maximum value of $a$ is $\frac{8}{3}$.