In triangle $ABC,$ $\overline{CD}$ is the bisector of angle $C,$ with $D$ on $\overline{AB}.$  If $\cos \frac{C}{2} = \frac{1}{3}$ and $CD = 6,$ compute
\[\frac{1}{BC} + \frac{1}{AC}.\]
The area of triangle $ABC$ is given by
\[[ABC] = \frac{1}{2} AC \cdot BC \cdot \sin C.\][asy]
unitsize (1 cm);

pair A, B, C, D;

A = (0,0);
B = (5,0);
C = (1,2);
D = extension(C, incenter(A,B,C), A, B);

draw(A--B--C--cycle);
draw(C--D);

label("$A$", A, SW);
label("$B$", B, SE);
label("$C$", C, N);
label("$D$", D, S);
label("$6$", (C + D)/2, NE);
[/asy]

We can also write
\begin{align*}
[ABC] &= [ACD] + [BCD] \\
&= \frac{1}{2} AC \cdot CD \sin \frac{C}{2} + \frac{1}{2} BC \cdot CD \sin \frac{C}{2} \\
&= 3AC \sin \frac{C}{2} + 3BC \sin \frac{C}{2} \\
&= 3 (AC + BC) \sin \frac{C}{2}.
\end{align*}Thus,
\[\frac{1}{2} AC \cdot BC \cdot \sin C = 3(AC + BC) \sin \frac{C}{2}.\]Then
\[AC \cdot BC \sin \frac{C}{2} \cos \frac{C}{2} = 3(AC + BC) \sin \frac{C}{2},\]so
\[\frac{AC \cdot BC}{3} = 3 (AC + BC).\]Hence,
\[\frac{1}{AC} + \frac{1}{BC} = \frac{AC + BC}{AC \cdot BC} = \boxed{\frac{1}{9}}.\]