O(n) as embedded submanifold I want to show that the set of orthogonal matrices, $O(n) = \{A \in M_{n \times n} | A^tA=Id\}$, is an embedded submanifold of the set of all $n \times n$ matrices $M_{n \times n}$. 
So far, I have used that this set can be described as $O(n) = f^{-1}(Id)$, where $f: M_{n \times n} \rightarrow Sym_n = \{A \in M_{n \times n} | A^t = A\}$ is given by $f(A) = AA^t$, and that the map $f$ is smooth. Hence I still need to show that $Id$ is a regular point of this map, i.e. that the differential map $f_*$ (or $df$ if you wish) has maximal rank in all points of $O(n)$. 
How do I find this map? I tried taking a path $\gamma = A + tX$ in $O(n)$ and finding the speed of $f \circ \gamma$ at $t=0$, which appears to be $XA^t + AX^t$, but don't see how to proceed. Another way I thought of was by expressing everything as vectors in $\mathbb{R}^{n^2}$ and $\mathbb{R}^{\frac{n(n+1)}{2}}$, but the expressions got too complicated and I lost track.

I think you have almost done. As you said, it suffices to show that $\mathrm{Id}$ is a regular value of $f$, i.e. for each $A\in O(n)$, $f_*:T_A M_{n\times n}\to T_{\mathrm{Id}}Sym_n$ is surjective, where $T_pX$ denotes the tangent space of $X$ at $p$. Note that $T_A M_{n\times n}$(resp. $T_{\mathrm{Id}}Sym_n$) can be identified with $M_{n\times n}$(resp. $Sym_n$) and, as you have known, $f_*(X)=XA^t+AX^t$. Then you only need to verify that for any $S\in Sym_n$, there exists $X\in M_{n\times n}$, such that $XA^t+AX^t=S$. At least you may choose  $X=\dfrac{1}{2}SA$.
