You can have a vertical asymptote where both the numerator and denominator are zero.
You don't always have an asymptote just because you have a $0/0$ expression.
Now if the denominator is zero (at the point in question) and the numerator is not zero, then you will have a vertical asymptote. This is exactly what it happening in the video where
$$
\lim_{x\to 3} f(x) = \lim_{x\to 0} \frac{9(x-9)}{x-3}
$$
This limit is $\pm\infty$ (depending on the side and so $x=3$ is an vertical asymptote.
Note also that in the example in the video, you have
$$
\lim_{x\to -3} f(x) = \lim_{x\to -3} \frac{3(x-9)}{x-3} = \frac{3(-12)}{-6}.
$$
Since this limit is not $\infty$ (or $-\infty$) the function does not have a vertical asymptote at $x=-3$.
Finding asymptotes is all about finding limits.