Answer
$9$ does not belong to the solution set
Work Step by Step
The mistake is in the second zero, which is not $9$.
$$y=(x+4)(x^2-9)$$
Factor the function completely:
$$y=(x-(-4))(x-3)(x+3)$$
$$y=(x-(-4))(x-3)(x-(-3))$$
Now it is in the form $y=a(x-p)(x-q)(x-z)$, so the zeros are $-4$ ,$3$ and $-3$.
The function has three zeros and $9$ is not among them.