Use Green's theorem to calculate the area enclosed by the curve:
$x^{2/3}+y^{2/3}=4$
Knowing that
$A=\frac{1}{2}\int_c xdy-ydx$
I know that there are already some questions and answers on this site regarding Green's theorem, and I've read many of them. I still really do not know how to proceed, though. Any help or direction would be appreciated. Thanks.
Alright, I am going to assume you are working over the reals, and make the assumption that $(-x)^{1/3} = -x^{1/3}$. (I.e., we are ignoring the existence of complex solutions.) Then the curve is symmetric about the $x$ and $y$ axes, so you need only compute the area in any of the four quadrants. (It's either this or evaluate the 4 smooth pieces in each quadrant. Either way works, but $3<4$, so why not.)
For this you want to parametrize each piece, and run through the standard line integral computation for each of the three (or four) pieces. Parametrizing the straight lines is not so bad, but for the curve you need to think a little more carefully. Firstly where does it start, and where does it end? (Not only what are the points, but you need to be careful with the choice of orientation for using Green's theorem properly!)
Next, how can we actually write down a parametrization? Well, you can simply set $x=t$, and find out what this forces $y$ to be. In particular, setting $x(t) = t$ will give you a well-defined function $y(t)$ so that the curve is plotted out by $(x(t),y(t))$. With this hand you can now do the standard line integral computation. (Not the nicest of integrals, but doable.)
Let me know if any of the steps are unclear.