If a first derivative doesn't exist at a certain point, is it not a critical point? Say $f'(x)$ of a function was $(x+2) \over (x+3)$. If $x = -2$ then $f'(x)$ = 0, which means that at this point, there would be a local min or max. But what if $x = -3$? It doesn't exist for $f'(x)$, so do we just ignore it? State it DNE at $x = -3$ so it is not a critical point?

I think the easiest explanation at a basic calculus level is to simply say that no, it is not a critical point. Hence if we want to do problems regarding optimization for a function which does not have its derivative (or the function itself) defined everywhere, then you would want to consider the "shape" of that function when deducing things, and not always just writing $f'(x)=0$. For example, if you want to minimise $1/x$, setting its derivative equal to $0$ you get $1=0$, which leads you to the conclusion that there are no critical points. Indeed, if you look at the graph, there aren't any "hills" or "valleys". Yet it still makes sense to say that its minimum value is arbitrarily large, or sometimes $-\infty$, in many contexts. 
