In triangle $ABC$, angle $C$ is a right angle and $CB > CA$. Point $D$ is located on $\overline{BC}$ so that angle $CAD$ is twice angle $DAB$. If $\frac{AC}{AD} = \frac{2}{3}$, then find $\frac{CD}{BD}.$
Let $a = \angle DAB = \frac{1}{2} \angle CAD.$  Since $\frac{AC}{AD} = \frac{2}{3},$ without loss of generality, we can assume $AC = 2$ and $AD = 3.$  Then by Pythagoras on right triangle $ACD,$ $CD = \sqrt{5}.$

[asy]
unitsize(2 cm);

real u = 5/9;
pair A, B, C, D, E;

A = (0,0);
C = (2*u,0);
D = (2*u,sqrt(5)*u);
E = interp(C,D,2/5);
B = extension(A, reflect(A,D)*(E), C, D);

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

label("$A$", A, SW);
label("$B$", B, NE);
label("$C$", C, SE);
label("$D$", D, dir(0));
label("$2$", (A + C)/2, S);
label("$3$", (A + D)/2, NW);
label("$\sqrt{5}$", (C + D)/2, dir(0));
[/asy]

From right triangle $ACD,$ $\cos 2a = \frac{2}{3}.$  Then from the half-angle formula,
\begin{align*}
\sin a &= \sqrt{\frac{1 - \cos 2a}{2}} = \frac{1}{\sqrt{6}}, \\
\cos a &= \sqrt{\frac{1 + \cos 2a}{2}} = \sqrt{\frac{5}{6}}.
\end{align*}By the Law of Sines on triangle $ABD,$
\[\frac{BD}{\sin a} = \frac{3}{\sin (90^\circ - 3a)},\]so
\[BD = \frac{3 \sin a}{\cos 3a} = \frac{3 \sin a}{4 \cos^3 a - 3 \cos a} = \frac{3 \cdot \frac{1}{\sqrt{6}}}{\frac{4 \cdot 5 \cdot \sqrt{5}}{6 \sqrt{6}} - \frac{3 \sqrt{5}}{\sqrt{6}}} = \frac{9}{\sqrt{5}}.\]Hence,
\[\frac{CD}{BD} = \frac{\sqrt{5}}{9/\sqrt{5}} = \boxed{\frac{5}{9}}.\]