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I want to prove that $M:=\{(x,y)\in\mathbb{R}^3\times\mathbb{R}^3:\|x\|=\|y\|=1, \langle x,y\rangle=0\}$ is diffeomorphic to $\text{SO}(3,\mathbb{R})$.

By now, I have proven via the regular value theorem that $M$ is a submanifold of $\mathbb{R}^3\times\mathbb{R}^3$, considering the morphism $\zeta:\mathbb{R}^3\times\mathbb{R}^3\rightarrow\mathbb{R}^3$ given by $$\zeta(x,y)=\left(\|x\|-1,\|y\|-1,\langle x,y\rangle\right)$$ where $M=\zeta^{-1}(0,0,0)$. How should I proceed now?

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    I'm not really clear on what your saying. You want to prove $M$ is diffeomorphic to $SO(3)$? If I understand your definition of $M$ I don't see how these are diffeomorphic, the dimension is off. Remember $SO(3)$ is diffeomorphic to $\mathbb{RP}^n$. – 2017-01-04
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    @Wintermute But $\text{SO}(3)$ has dimension 3 just like $M$, I don't see any problem with the dimension. This can be seen via $F:\mathcal{M}(3,\mathbb{R})\rightarrow\text{S}(3,\mathbb{R})\simeq\mathbb{R}^6$ defined as $F(A)=A^TA$ – 2017-01-04
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    The proof is quite simple you already have two columns of an orthogonal matrix. Note, you get the third column vector by $z=x\times y$. – 2017-01-04

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