9. (10 points) As shown in the figure, a cube $X Y T Z-A B C D$ is cut along two planes $B C T X$ and $B D T Y$, dividing the cube into 4 pieces. What fraction of the cube's volume is the piece that contains vertex $A$?
【Analysis】Cut along the plane $B C T X$, at this point the piece containing vertex $A$ accounts for $\frac{1}{2}$ of the volume of the cube, then cut along $B D T Y$, the piece containing vertex $A$ accounts for $\frac{2}{3}$ of the piece cut along the plane $B C T X$, by the multiplication principle, the piece containing vertex $A$ accounts for the volume of the cube as: $\frac{1}{2} \times \frac{2}{3}=\frac{1}{3}$.
【Solution】Solution: Cut along the plane $B C T X$, at this point the piece containing vertex $A$ accounts for $\frac{1}{2}$ of the volume of the cube, then cut along $B D T Y$, the piece containing vertex $A$ accounts for $\frac{2}{3}$ of the piece cut along the plane $B C T X$, so the piece containing vertex $A$ accounts for the volume of the cube as: $\frac{1}{2} \times \frac{2}{3}=\frac{1}{3}$. Answer: The piece containing vertex $A$ accounts for $\frac{1}{3}$ of the volume of the cube.