Yes, 1 divided by g(x).  g(x) on its own is, well, g(x).

Let's just carry on and see how this plays out. lolz.  I confirmed with a TA in the area that I'm studying that I don't have to treat it as a 1/3 if I'm only working with g(x).  Let's carry on!

From the first derivative,

f(x) = 2x
f'(x) = 2

g(x) = 3(x^2-1)^(2/3)
g' = (2/3)(2x)(x^2-1)^(-1/3)

So now, we use the quotient rule:

(f'g - fg')
-------
    g^2

(2*3(x^2-1)^(2/3)) - ((2x)((2/3)(2x)(x^2-1)^(-1/3)))
---------------------------------------------
                  ( 3(x^2-1)^(2/3))^2

That's what I believe it should be (not fully simplified, but screw it. :P)