Consider a smooth manifold $M = \{ (u,v) \in \mathbb{R^3} \times \mathbb{R^3} \mid \|u\|=\|v\|=1 \text{ with } u \perp v \}$, and want to show $M$ is diffeomorphic to $SO(3)$, the rotational group in $\mathbb{R^3}$.
I first define a map $f:M \rightarrow SO(3)$ by $(u,v) \mapsto [u,v,u \times v]$
How do I show $f$ is smooth with smooth inverse $f^{-1}$? And this is the only way to show diffeomorphism right?
You're right: every element in $SO(3)$ is just a positively oriented basis of $\mathbb R^3$, and such a basis is determined by its two first vectors as you propose. In other words, one represents motions in $SO(3)$ by their orthogonal matrices with respect to the standard basis $\{e_1,e_2,e_3\}$. Now $M\subset\mathbb S^2\times\mathbb S^2$ is defined by $h(u,v)=\langle u,v \rangle=0$ and $h:\mathbb S^2\times\mathbb S^2\to\mathbb R$ is a submersion: $$ d_{(u,v)}h(v,0)=\langle v,v\rangle=\|v\|^2\ne0. $$ for any $(u,v)\in M$ (because $v\in T_u\mathbb S^2$). The map $f(u,v)=[u,v,u\times v]$ is smooth, because $L=[u,v,u\times v]$ is just the matrix of the motion and vector products are algebraic operations with the coordinates of $u,v$. The inverse is given by $u=L(e_1), v=L(e_2)$, again smooth. Hope this helps!