Maximizing Differentiable Function

Using the Mean Value Theorem , there exists some value $x$ in the interval $(3,9)$ such that $f'(x)=\frac{f(9)-f(3)}{9-3}=\frac{f(9)-10}{6}$ Since $f'(x)\le 9$ , $\frac{f(9)-10}{6}\le 9$ $\implies f(9)\le 9\cdot 6+10=64$ Therefore, the maximum possible value of $f(9)$ is $\boxed{64}$ .

Note: Consider the function $f(x)=9x-17$ . The function is differentiable on $[3,9]$ , $f(3)=10$ and $f'(x)=9$ . Therefore, this function satisfies all the given conditions. Also, $f(9)=64$ which shows that the maximum value can be achieved.