Let $F: \mathbb{R}^2 \rightarrow \mathbb{R}^3$ be defined via
$$(u,v) \mapsto (u,v,u^2-v^2)$$
And let $\omega$ be the $2$-form $\omega = y dx \wedge dz + x dy \wedge dz$
Compute the pull back $F^*\omega$. They have
\begin{align} F^*(y dx \wedge dz + x dy \wedge dz)&= v du \wedge d(u^2-v^2)+udv \wedge d(u^2-v^2)\\ &= v du \wedge (2udu-2vdv)+ udv \wedge (2udu-2vdv). \end{align}
Does the last line equal to:
$$2uv (du \wedge du) - 2v^2 du \wedge dv +2u^2 (dv \wedge du)-2vu(dv \wedge dv).$$
Is what I have above ^^ correct?
And since $du \wedge du=0$ and $(du \wedge dv)=-(dv \wedge dv)$ this reduces to
$$F^*\omega = -2(u^2+v^2)du \wedge dv.$$