I embed the 2-sphere and the 3-sphere in $\mathbb{R}^3$ and $\mathbb{R}^4$ respectively. Then denote by $\{x_1,x_2,x_3,x_4\}$ the coordinates on the 3-sphere and $\{y_1,y_2,y_3\}$ on the 2-sphere. So $\sum_{i=1}^4 x_i^2 = 1 = \sum_{i=1}^3 y_i^2$.
Now consider the 2-forms $\omega=dx_1\wedge dx_2+dx_3\wedge dx_4$ on $S^3$ and $\alpha=y_1 dy_2\wedge dy_3+y_2 dy_3\wedge dy_1 + y_3 dy_1\wedge dy_2$ on $S^2$.
I wish to show that when pulling back $\alpha$ to $S^3$ using the Hopf map $H:S^3\rightarrow S^2$ given by
$(x_1,x_2,x_3,x_4)\mapsto \left(x_1^2+x_2^2-x_3^2-x_4^2, 2(x_2x_3 - x_1 x_4), 2(x_2x_4+x_1x_3)\right)$,
the result is proportional to $\omega$, so:
$H^*\alpha=c\omega$ for some $c\in\mathbb{R}$.
I tried starting from the left hand side, so by definition of the pullback I just wrote down the form $\alpha$ with the $y_i$ replaced by the $i^{th}$ component of $H(x_1,x_2,x_3,x_4)$, and then I used $\sum_{i=1}^4 x_i^2 = 1$ and $\sum_{i=1}^4 x_idx_i = 0$. This however became a huge mess and nearly impossible to do by hand...
Is there an easier way? Is there a way to take $H$ to the right hand side, using some properties of the pull back? Then I could start from the right hand side, which might be easier.
Your 2-form on the 2-sphere is the area form, while the 2-form on the 3-sphere is the contact form. The point is that both are invariant under a suitable group of transformations. In the case of the 2-sphere it is SO(3) (at least), whereas in the case of the 3-sphere it is the restriction of SU(2) (at least). Both of these act transitively. Use this to reduce the general calculation to the calculation at a single point, say $(0,0,1)$ of the 2-sphere.