Answer: 7

$9^9=\left(3^2\right)^9=3^{18}$

The factors of $3^{18}$ are $1,3,3^2,3^3,...,3^{17},3^{18}$. (There are no others since $3$ is prime.)

Cubes: The factors where the power is a multiple of $3$ are cubes, for example $3^{15} = 3^{3\times5} = \left(3^5\right)^3$.

The factors where the power is not a multiple of $3$ are not cubes, because $3$ is prime, so there is no other way to divide the product into $3$ equal parts.

This means the cube factors are $1, 3^3, 3^6, 3^9, 3^{12}, 3^{15}, 3^{18}$.

Therefore there are $7$ cube factors of $9^9$.