9. Given the function $f(x)=a x^{3}+b x^{2}+c x+d(a \neq 0)$, when $0 \leqslant x \leqslant 1$, $\left|f^{\prime}(x)\right| \leqslant 1$, try to find the maximum value of $a$.
$$
\begin{array}{l}
f^{\prime}(x)=3 a x^{2}+2 b x+c \Rightarrow\left\{\begin{array}{l}
f^{\prime}(0)=c, \\
f^{\prime}(1)=3 a+2 b+c, \\
f^{\prime}\left(\frac{1}{2}\right)=\frac{3}{4} a+b+c
\end{array}\right. \\
\Rightarrow a=\frac{2}{3}\left[f^{\prime}(0)+f^{\prime}(1)-2 f^{\prime}\left(\frac{1}{2}\right)\right] \leqslant \frac{2}{3}\left[\left|f^{\prime}(0)\right|+\left|f^{\prime}(1)\right|+2\left|f^{\prime}\left(\frac{1}{2}\right)\right|\right] \leqslant \frac{8}{3} . \\
\text { When the equality holds, }\left\{\begin{array} { l } 
{ f ^ { \prime } ( 0 ) = c = 1 , } \\
{ f ^ { \prime } ( 1 ) = 3 a + 2 b + c = 1 , } \\
{ f ^ { \prime } ( \frac { 1 } { 2 } ) = \frac { 3 } { 4 } a + b + c = - 1 }
\end{array} \quad \Rightarrow \left\{\begin{array}{l}
a=\frac{8}{3}, \\
b=-4, \\
c=1 .
\end{array}\right.\right.
\end{array}
$$