First of all, I'm new to multivariable calculus... in a multivariable function, by assuming that its domain is going to be $R^{2}$ and its image is going to be all real numbers, the graph of that function is defined as a subset of $R^{3}$ in which the $x$ and $y$ axis are going to receive the inputs, and the output is going to be in $z$ $(x,y,f(x,y))$ ? Is that correct? Will its graph, in this example, be some kind of surface?
Thanks.
Yes. If such a $f(x,y)$ is defined for all real pairs $(x,y)$, for each such pair corresponds a single real value, which could be taken as the $z$ coordinate of a point. This interpretation leads to a surface, as you suppose.
Using Wolfram Alpha, for instance, you can get the plots of the following functions. Guessing the surfaces before looking them up would be an interesting exercise.
$f(x,y)=x$
$f(x,y)=y$
$f(x,y)=x+y$
$f(x,y)=x-y$
After you understand these graphs, you may look into surfaces such as this one which are still quite simply expressed, but fairly more interesting.