6. As shown in Figure 2, given that $N$ is a point on the semicircle with diameter $AB$, $NM \perp AB$ at point $M$, semicircles $\odot O_{1}$ and $\odot O_{2}$ have diameters $AM$ and $BM$ respectively, $\odot O_{3}$ and $\odot O_{4}$ are both internally tangent to the semicircle $\odot O$, also tangent to $MN$, and externally tangent to semicircles $\odot O_{1}$ and $\odot O_{2}$ respectively. If the sum of the areas of $\odot O_{3}$ and $\odot O_{4}$ is $64 \pi$, then the value of $\frac{1}{AM} + \frac{1}{BM}$ is ( ).
(A) 1
(B) $\frac{\sqrt{2}}{4}$
(C) $\frac{\sqrt{2}}{8}$
(D) $\frac{\sqrt{2}}{16}$
6. D.

As shown in Figure 5, let the radii of $\odot O_{1}, \odot O_{2}, \odot O_{3}, \odot O_{4}, \odot O$ be 
$R_{1}, R_{2}, R_{3}, R_{4},$ and 
$R$, respectively. $\odot O_{4}$ is tangent to $MN$ at point $D$. Draw $O_{4}C \perp AB$ at point $C$, and connect 
$$
OO_{4}, O_{2}O_{4}, DO_{4}.
$$

It is easy to see that quadrilateral $CO_{4}DM$ is a rectangle. Therefore, $CM = DO_{4} = R_{4}$. Hence, 
$$
OC = OO_{2} - CO_{2}
$$
$$
\begin{array}{l}
= (R - R_{2}) - (R_{2} - R_{4}) \\
= R - 2R_{2} + R_{4}.
\end{array}
$$

In right triangles $\triangle OCO_{4}$ and $\triangle O_{2}CO_{4}$, 
$$
OO_{4}^{2} - OC^{2} = CO_{4}^{2} = O_{2}O_{4}^{2} - CO_{2}^{2},
$$

which means
$$
\begin{array}{l}
(R - R_{4})^{2} - (R - 2R_{2} + R_{4})^{2} \\
= (R_{2} + R_{4})^{2} - (R_{2} - R_{4})^{2}.
\end{array}
$$

Thus, $R_{4} = R_{2} - \frac{R_{2}^{2}}{R} = R_{2} - \frac{R_{2}^{2}}{R_{1} + R_{2}} = \frac{R_{1} R_{2}}{R_{1} + R_{2}}$. Similarly, $R_{3} = \frac{R_{1} R_{2}}{R_{1} + R_{2}} = R_{4}$. Given that
$$
\pi R_{3}^{2} + \pi R_{4}^{2} = 64 \pi,
$$

we have $R_{3} = 4 \sqrt{2}$. Therefore,
$$
\frac{1}{AM} + \frac{1}{BM} = \frac{1}{2R_{1}} + \frac{1}{2R_{2}}
$$
$$
= \frac{R_{1} + R_{2}}{2R_{1}R_{2}} = \frac{1}{2R_{3}} = \frac{1}{8\sqrt{2}} = \frac{\sqrt{2}}{16}.
$$