When learning about rational equations, we learn that we must consider the domain and/or denominator, to ensure that our solution does not cause a division by 0 error. For example:
$$ \frac{x + 1}{9 - x} = \frac{2}{3} $$
Here, we say that $x$ has a solution but can never be 9.
But if we manipulate the equation with cross-multiplication, we get:
$$ x + 1 = \frac{2}{3}\left(9-x \right) $$
And in this form, we do not have the possibility of division by 0.
Are these two equations not equivalent? In both forms, $x = 3$. As such, why do we care about ensuring that, in the first form, we are mindful of the domain $x \neq 9$?
When dealing with equations, although it is common practice to perform operations such as multiplication and division to both sides of an equation, we need to be mindful that such operations can introduce or eliminate solutions. In the example that you gave, you were lucky. The rational equation that you started with, and the linear equation you ended with are equivalent since they have the same solution set $x=3$ as you observed. However, this won't always be the case.
For example, if we consider $\frac{x^2-1}{x+1} = 0$ and $x^2-1 = 0$, we have a rational and quadratic equation, respectively. It is easy to get from one equation to the other. If we start from the first equation with $\frac{x^2-1}{x+1} = 0$, we have to note that $x \neq -1$. If we go about trying to solve the rational equation, we will eventually solve $x^2-1 = 0$ and learn that $x = 1$ and $x = -1$ are potential solutions. However, now we have to remember that we weren't originally asked to solve $x^2-1 = 0$. We originally wanted to solve $\frac{x^2-1}{x+1} = 0$. Here the restriction that $x \neq -1$ is useful because it helps us eliminate $x = -1$ as a possible solution and conclude that only $x = 1$ is a solution. Only after we have performed a little bit of work can we see that both $\frac{x^2-1}{x+1} = 0$ and $x^2 - 1 =0$ have different solutions and as such are not equivalent.
So, to answer your questions "Are these two equations not equivalent?" - the two equations you gave have the same solutions, so there are equivalent. But as we have seen with the example I gave, that won't always be the case.
"As such, why do we care about ensuring that, in the first form, we are mindful of the domain $x \neq 9$?" - because a priori, you don't know whether or not $x = 9$ will be a solution to the equation $x + 1 = \frac{2}{3}(9-x)$. Only after you do a little bit of work simplifying the equation do you find out the solution. We give ourselves the restriction $x \neq 9$ at the beginning just in case $x = 9$ pops up as a potential solution during the course of our work.