Suppose I wanted to compute $f_{X}(x)$ or $f_{Y}(y)$ for $f(x, y) = 24xy$ where $0 < x < 1 $, $0 < y < 1$ and $0 < x + y < 1$.
I'm not sure about how to deal with the last constraint. I know that if the last constraint wasn't there, then I'd have
$$f_{X}(x) = \int_{0}^{1} f(x, y) \mathop{dy} $$
and
$$f_{Y}(y) = \int_{0}^{1} f(x, y) \mathop{dx}.$$
But, if we have $0 < x + y < 1$, is there anything different? Is it even possible?
This is a problem that I'm coming up by myself, so maybe it isn't even a possible answer, I was just wondering if it would be possible to calculate $f_{X}$ and $f_{Y}$ in this instance.
You cannot just ignore one constraint while finding the marginal densities.
Observe that the joint pdf $f$ of $(X,Y)$ is defined over the region $$S=\{(x,y):0<x,y<1,0<x+y<1\}$$
$S$ is the support of $(X,Y)$, and everywhere outside $S$, $f$ takes the value zero.
That is, the joint density of $(X,Y)$ should be written like
$$f(x,y)=24xy\,\mathbf1_{x,y\,\in \,S}$$
, where $\mathbf 1_{x\in A}$ is a shorthand for $$\mathbf1_{x\in A}=\begin{cases}1&,\text{ if }x\in A\\0&,\text{ otherwise }\end{cases}$$
Marginal pdf of $X$ should be a function of $x$, so you have to integrate over all such $y$ such that $0<y<1$ and $0<x+y<1$ holds simultaneously, i.e. when $0<y<1-x$.
Marginal pdf of $X$ is thus
$$f_X(x)=\int_0^{1-x} f(x,y)\,dy\,\mathbf1_{0<x<1}$$
Proceed similarly for the marginal pdf of $Y$.