In the unit cube $A B C D-A_{1} B_{1} C_{1} D_{1}$, point $E$ is on edge $A \Lambda_{1}$, and $A E=\frac{3-\sqrt{6}}{3}$. A plane $\alpha$ is constructed through points $E, C, D_{1}$. The dihedral angle between $\alpha$ and the face $D C C_{1} D_{1}$ is $\qquad$
9. $60^{\circ}$.
$\alpha k_{j}$ face $I C C_{1} D_{1}$ has two common points, therefore line $C D_{1}, \alpha$ intersects face $\triangle B B_{1} A_{1}$ at point $E$, thus forming intersection line $E F$, and $E F / / C D_{1}$, forming trapezoid $E F C D_{1}$ (as shown in the figure).

Draw a line through $E$ parallel to $A D$, intersecting $D D_{1}$ at $H$, then $E H$ lies in face $D C C_{1} D_{1}$, and $E H=1$. In square $D C C_{1} D_{1}$, draw $H P \perp C D_{1}$ at $P$, connect $E P$, by the theorem of three perpendiculars, we get $E P \perp C D_{1}$. Therefore, $\angle E P H$ is the half-angle of the intersection angle between $\alpha$ and face $D C C_{1} D_{1}$.

Given $D_{1} H - A_{1} E = 1 - A E = \frac{\sqrt{6}}{3}$, we know $H P = \frac{1}{\sqrt{2}} D_{1} H = \frac{\sqrt{3}}{3}$. Thus, $\tan \angle E P H = \frac{E H}{H P} = \sqrt{3}$. Therefore, $\angle E P H = 60^{\circ}$.