Does simultaneously finding max/min in array of coordinates by x and y increase comparisons?
does this double the amount of comparisons or can I still claim it’s $O(3n/2)$?


You misunderstand the big $O$ notation. 

Even if you double or quadruple the amount of comparisons, you can still claim it is $O(3n/2)$. That is, the amount of comparisons as a function of $n$ is asymptotically not bigger than $3n/2$.  The very purpose of big $O$ notation is to hide a constant factor so that it becomes clearer what is the asymptotic picture. In fact, $O(3n/2)$ = $O(n)$. When I write "=", I mean equality. That is, the left hand side and the right hand side represent exactly the same set of functions.

Here is what you could say.

If I use this method to find the min/max for the left/right box boundaries and then use the same algorithm to find min/max of y for the top/bottom boundaries, I will use $2(\lceil 3n / 2\rceil - 2) \le 3n -3$ comparisons. That is, I will use $O(n)$ comparisons.