Intuition or motivation for the definition of an hypersurface. What are we actually trying to define? 
If we have $x^2 + y^2= 1 $ then we can solve for $y$ and $x$, at least in parts. The implicit function theorem gives us the conditions to solve these things. At this part of this book (Folland, about PDE), it says that if a hypersurface is given by that definition, we can use the implicit funciton theorem to solve for some variable in relation to the other $n-1$ other ones.
So how does this definiton arise? I think the open sets $S$ and $V$ have something to do with locality of the solution, as we have inverse solutions for $x^2 + y^2 = 1$ only locally. 
The definition makes no sense to me, I'd like some help.
UPDATE:
It also talks about hyperplane without defining it. The only definition is of an hypersurface, which is also something I don't unerstand. What is one and another?
Also, the implicit funciton theorem says that the jacobian must be invertible. The closest the book talks about this is when it says the gradient is nonvanishing, but I don't think it implies jacobian invertible.
UPDATE:
I'm trying to visualize it. The plane should be the subset $S$ of $\mathbb{R}^3$. $V$ is an open around $x_0$. If for every $x_0$ and every open $V$ we can find a function $\phi\in C^k(V)$ with $\nabla \phi$ nonvanishing on $S\cap V $ and $S\cap V  = \{x\in V: \phi(x) = 0\}$, then the plane is an hypersurface. 
There's no motivation anywhere on google (there's so little about the word hypersurface in general). So the main question is: what is this crazy definition suppose to define? What are the challenges in defining an hypersurface? (what even is an hypersurface?).
Please remember that you're explaining to someone who have little background on this manifold thing etc (actually I don't even know what this term means at all) so a little background explanation would be good. 
 A: Idea: A surface in $\mathbb{R}^3$ is something two-dimensional, and a hypersurface in $\mathbb{R}^n$ is something $(n-1)$-dimensional (one dimension less than the surrounding space).
But this is too vague to be a real definition, so we must try to make it more precise.
A typical thing that we think of as a (smooth) surface in $\mathbb{R}^3$ is the graph of a (smooth) function, $z = f(x,y)$. So in $\mathbb{R}^n$, the graph of a nice function
$$
x_n = f(x_1,x_2,\dots,x_{n-1})
$$
ought to count as an example of a hypersurface. It's $(n-1)$-dimensional, since we can vary the $n-1$ quantities $x_1, \dots, x_{n-1}$ however we like, and the $n$th variable $x_n$ will adapt its value accordingly.
But that's not the only kind of hypersurface that we can imagine. For example, a graph of the form
$$
x_1 = f(x_2,\dots,x_{n-1},x_n)
$$
should do just as well.
This leads to the idea of defining a hypersurface as something that locally (in an open set around each point on the surface) has the form “one of the variables is a smooth function of the other $n-1$ variables”.
And this is where the implicit function theorem enters. It says that the zero set of a smooth function $\phi\colon \mathbb{R}^n \to \mathbb{R}$ has precisely this form, provided that the gradient $\nabla \phi$ is nonzero at each point of the set. (If $\nabla \phi(P)$ is not the zero vector, then at least one partial derivative $\partial \phi/\partial x_k(P)$ must be nonzero, and then the implicit function theorem says the equation $\phi=0$ locally around the point $P$ defines $x_k$ as a smooth function of the other variables.)
(When you say that the Jacobian should be invertible, I think you are confusing the implicit function theorem with the inverse function theorem.)
So that's what a hypersurface is. A hyperplane is simply the solution set of a linear equation
$$
C_1 x_1 + \dots C_n x_n = D
$$
where not all $C_k$ are zero. (It's an elementary fact from linear algebra that you then get a solution with $n-1$ parameters.) And this is of course a special case of a hypersurface, since if $C_k \neq 0$ you can easily solve for $x_k$ in terms of the other variables.
