9. Given that the line $l$ with slope $k$ passing through the point $P(3,0)$ intersects the right branch of the hyperbola $C: x^{2}-\frac{y^{2}}{3}=1$ at points $A$ and $B$, and $F$ is the right focus of the hyperbola $C$, and $|A F|+|B F|=16$, find the value of $k$.
Let $AB: y=k(x-3)$, combining $\left\{\begin{array}{l}y=k(x-3), \\ 3 x^{2}-y^{2}=3\end{array} \Rightarrow 3 x^{2}-k^{2}(x-3)^{2}=3\right.$ $\Rightarrow\left(3-k^{2}\right) x^{2}+6 k^{2} x-9 k^{2}-3=0 \Rightarrow x_{1}+x_{2}=\frac{6 k^{2}}{k^{2}-3}$.
Since $|AF|=e x_{1}-a,|BF|=e x_{2}-a \Rightarrow|AF|+|BF|=e\left(x_{1}+x_{2}\right)-2 a=\frac{12 k^{2}}{k^{2}-3}-2=16$ $\Rightarrow \frac{2 k^{2}}{k^{2}-3}=3 \Rightarrow 2 k^{2}=3 k^{2}-9 \Rightarrow k= \pm 3$. At this point, $k^{2}>3$, satisfying the condition. Therefore, the value of $k$ is $\pm 3$.