I have the following density function. $$f(x,y)=\frac{1}{32}*(10-3x^2-y)\quad,-1<x<1; 0<y<2$$ When calculating the marginal density for $x$, $f_x$, I get $$\int_{0}^2\frac{1}{32}*(10-3x^2-y) dy \\ = [10y-3x^2y-\frac{y^2}{2}]^2_0 \\ \Rightarrow f_x = \frac{9}{16}-\frac{3x^2}{16}$$
I am wondering however over which domain this function is defined. Where do I start thinking/Why?
It is common to think of PDF's as measurable non-negative functions $\mathbb R^n\to\mathbb R$ and in that context the domain of a PDF is just $\mathbb R^n$.
Quite often there is a situation that a random variable $X$ only takes values on e.g. some interval $(a,b)$ and in that situation on forehand we can choose for a corresponding PDF that takes value $0$ outside $(a,b)$.
In your case I would say that the domain of $f_X$ is $\mathbb R$ and that it is prescribed by $x\mapsto\frac9{16}-\frac{3x^2}{16}$ for $x\in(-1,1)$ and by $x\mapsto0$ otherwise.
Observe that for a fixed $x\notin(-1,1)$ indeed you get: $$f_X(x)=\int_{-\infty}^{\infty} f(x,y)dy=0$$ if $f(x,y)$ is prescribed by $(x,y)\mapsto0$ for $x\notin(-1,1)$.