Algebraic surface as a smooth manifold

Question

Let $S$ be the set of points $x=(x_1,x_2,\ldots,x_9)\in \mathbb{R}^9$ which satisfy the following conditions:
$$x_1^{2}+x_2^{2}+x_3^{2}=x_4^{2}+x_5^{2}+x_6^{2}=x_7^{2}+x_8^{2}+x_9^{2}=1$$ $$x_1x_4+x_2x_5+x_3x_6=x_1x_7+x_2x_8+x_3x_9=x_4x_7+x_5x_8+x_6x_9=0,$$ Thus we have a compact subspace (really, it is closed and bounded) in $\mathbb{R}^9$. How to prove it has a structure of a $3$-manifold?
I understand, that it is necessary to take into account some special properties of these equations. That's why I want to know whether there are some general theorems that link such kind of surfaces with smooth structures.
I will be grateful for some hints to show $S$ is a smooth $3$-manifold.

Answer 1

Hint: $F^{-1}(a)$ with $a \in \mathbb{R}^n$ is an $m - r$ manifold (where $r$ is the rank of the Jacobian of $F$) if and only if the Jacobian of $F$ has maximum rank in $a$. ($F\colon \mathbb{R}^m \to \mathbb{R}^n$)

EDIT: This is actually more general, since it works for smooth maps between smooth manifolds. Let $F\colon M \to N$ be a smooth map of smooth manifolds and let $a \in N$ be a regular value (that is the rank of the Jacobian of $F$ has maximum rank), then $F^{-1}(a)$ is a regular submanifold of $M$ with dimension $m -r$.

Of course for this to make sense one would have to know what a regular submanifold and the Jacobian of a map of manifolds are.

Answer 2

Because these defining functions are all homogeneous, you will find that Euler's Theorem on homogeneous functions will be helpful in checking the rank condition.