Nowhere $0$ form on the sphere? Consider the differential form on $\mathbb R^3$ given by $ x dy \wedge dz + y dz \wedge dx + z dx \wedge dy$. I converted this to spherical coordinates using a laborious calculation, and when I'm done, by some miracle (which would be cool if someone could explain exactly how that works), I get something really compact and elegant: $\rho^3 \sin \phi d\phi \wedge d\theta$. 
If we call this form $\Omega$, my task is to compute $i^*\Omega,$ where $i$ represents the inclusion from $\mathbb S^2 \rightarrow \mathbb R^3$, which seems to be pretty easy also - on the unit sphere we just have $\rho = 1$.
Now why is this nowhere $0$ on the sphere? It looks like it's $0$ along the half plane where $\phi = 0$!

Let me remark that there is no need to convert to spherical coordinates. The coordinates $x,y,z$ are well defined smooth functions on $S^2$, so their exterior derivatives are well defined one-forms on $S^2$. Naturality of the exterior derivative and the operations on differential forms shows that the pullback of your form to $S^2$ is still given by $xdy\wedge dz+ydz\wedge dx+zdx\wedge dy$. To show that this is nowhere vanishing, observe that on $S^2$, one has $x^2+y^2+z^2=1$, so applying the exterior derivative (and dividing by 2) we get $0=xdx+ydy+zdz$. Now on the open subset $\{x\neq 0\}$ you can use this equation to rewrite your form as $\frac1x(x^2+y^2+z^2)dy\wedge dz$. At the same time, on this open subset you can use $y$ and $z$ as local coordinates, which shows that $dy\wedge dz$ and hence your form is nowhere vanishing on $\{x\neq 0\}$. Similar arguments apply to the sets $\{y\neq 0\}$ and $\{z\neq 0\}$, and together the three sets cover $S^2$. 
