9. Given a cube $A B C D-A_{1} B_{1} C_{1} D_{1}$ with edge length 1, a sphere is constructed with vertex $A$ as the center and $\frac{2 \sqrt{3}}{3}$ as the radius. The length of the curve formed by the intersection of the sphere's surface with the cube's surface is equal to
9. $\frac{5 \sqrt{3} \pi}{6}$.

As shown in Figure 6, the sphere intersects with all six faces of the cube, resulting in two types of intersection lines: one type on the three faces containing vertex $A$; the other type on the three faces not containing vertex $A$.

On the face $A A_{1} B_{1} B$, the intersection line is $\overparen{E F}$ and lies on the great circle passing through the center of the sphere $A$.

Since $A E=\frac{2 \sqrt{3}}{3}$, $A A_{1}=1$, we have
$\angle A_{1} A E=\frac{\pi}{6}$.
Similarly, $\angle B A F=\frac{\pi}{6}$.
Therefore, $\angle E A F=\frac{\pi}{6}$.
Thus, the length of $\overparen{E F}$ is $\frac{2 \sqrt{3}}{3} \times \frac{\pi}{6}=\frac{\sqrt{3} \pi}{9}$. There are 3 such arcs.
On the face $B B_{1} C_{1} C$, the intersection line is $\overparen{F G}$ and lies on the small circle formed by the intersection of the plane at a distance of 1 from the center of the sphere with the sphere, with the center of the small circle at $B$, and the radius $\frac{\sqrt{3}}{3}$, $\angle F B G=\frac{\pi}{2}$.
Thus, the length of $\overparen{F G}$ is $\frac{\sqrt{3}}{3} \times \frac{\pi}{2}=\frac{\sqrt{3} \pi}{6}$. There are also 3 such arcs. Therefore, the total length of the curves is
$$
3 \times \frac{\sqrt{3} \pi}{9}+3 \times \frac{\sqrt{3} \pi}{6}=\frac{5 \sqrt{3} \pi}{6} .
$$