Show that $\int_{S^2} \omega = 4\pi$ 
Here $$\omega = \dfrac{x \ dy \wedge dz + y \ dz \wedge dx + z\ dx \wedge dy}{(x^2+y^2+z^2)^{3/2}}$$.
I have calculated $f^\star \omega = \sin(u) \ du \wedge dv$ which I believe is correct. Could someone help with part $d$ I have no idea how it relates to part c!
 A: This will probably only be of limited help, and is probably too late to be of any help, but...
What is your reference text? From the question, it looks like that by definition the integral depends on the parametrisation. Namely, $S^2$ denotes the unit sphere together with the parametrization $f(u,v) = (\,x(u,v),\ y(u,v),\ z(u,v)\,)$, and  the integral is a function on $f$: $f \mapsto \int_{S^2} \omega = \int_f \omega$. So - taking Rudin, which takes this point of view, as reference -  by definition $$\int_{S^2} \omega =\int_{u=0}^\pi \left( \int_{v=0}^{2\pi}\, \sin u \,dv \right)\,du$$ (sections 10.11 and 10.1 - the former for differential forms and parametristations, and the latter for the definition of the integral in $\mathbb R^n$). 
This corresponds to c), because, by the change of variables formulism, $\int_{S^2} = \int_{Id} f^*\omega$, where $Id:[0,\pi]\times[0,2\pi] \to [0,\pi]\times[0,2\pi]$ is the identity (theorem 10.24). 
Edit: in fact, in the first version of this, I was tempted to insert the pull-back into my "by definition" phrase/equality :
$$\int_{S^2} \omega =\int _{[0,\pi]\times [0,2\pi]} f^*\omega = \int_{u=0}^\pi \left( \int_{v=0}^{2\pi}\, \sin u \,dv \right)\,du,$$
but did not because Rudin does not introduce the pull-back (with different notation) until later; instead, Rudin writes out explicitly the Jacobian formula to define the integral of a form as an integral on a certain type of set in $\mathbb R^n$ (as I did above). But the explicit Jacobian formula is the pull-back... 
Note: Rudin allows parametrisation domains of the form $D = [a_1,b_1]\times \cdots [a_n,b_n]$, $a_k < b_k$, whereas by your comments to Will P., it looks like you only allow $D=[0,1]^n$. So I actually would have preferred writing this as a comment, rather than as an answer, but it's too long. 
A: If you insert what you got for w in the integral, obviously, there is not much to go with. the solution is to consider how you integrate over the sphere - as a triple integral in x,y,z coordinates, the integral limits are pretty complicated, as the limit for each axis depends on the value of the others.
The trick is to change coordinate system, and transfer the integral over S^2 into an double integral of u du and v dv, where one of them goes from -pi to pi, and the other from 0 to 2*pi. Now you transform your term for w (as shown above) to u and v, and then the integral is actually very easy to execute.
i'm not gonna do the math for you, as 1. I don't know how to make formulas here, and 2. its your homework. I had to do the exact same thing thirty years ago... I'm sure it still gives 4 * pi.
A: d) On the unit sphere $x^2+y^2+z^2=1$; then $w$ is simply the surface area form (as in 
Volume form on $(n-1)$-sphere $S^{n-1}$); integrating it over the whole sphere gives the area, $4\pi$.
c) In other words it's asking you to write the area form in spherical coordinates, which you did correctly.
