Implicit function theorem and implicit differentiation This is perhaps something simple; but I am not quite getting why the implication is true; I seem to be missing something.
Supposedly, the implicit function theorem:

Let $f: \mathbb{R}^{n + m} \rightarrow \mathbb{R}^m$ be a continuously differentiable function, and let $\mathbb{R}^{n+m} $ have coordinates $( x, y)$, where $x \in \mathbb{R}^n$ and $y\in \mathbb{R}^m$. Fix a point $( a , b) = (a_1 , \ldots , a_n , b_1 , \ldots, b_m )$ with $ f( a, b) = c$, where $c \in \mathbb{R}^m$. If the matrix $( \partial f_i/\partial y_j)(a,b)$ is invertible,  then there exists an open set $U$ containing $a$, and an open set $V$ contntaining $b$, and a unique continuously differentiable function $g: U \rightarrow V$ such that 
  $$ \{ (\mathbf{x}, g(\mathbf{x}))|\mathbf x \in U  \} = \{ (\mathbf{x}, \mathbf{y}) \in U \times V| f(\mathbf{x}, \mathbf{y}) = \mathbf{c} \}.$$

implies that implicit differentiation is okay: 
$$ \frac{dy}{dx} = -\frac{\partial F / \partial x}{\partial F / \partial y}.$$
What am I missing here? I again apologize if this is something very trivial.
 A: Let $ F: \mathbb{R}^{2} \to \mathbb{R} $ be a continuously differentiable function. Fix a point $ (a,b) \in \mathbb{R}^{2} $, and let $ c = F(a,b) $. Next, compute the Jacobian of $ F $:
$$
\forall (x,y) \in \mathbb{R}^{2}: \quad [\mathbf{D}(F)](x,y) =
\left[ \matrix{{\partial_{1} F}(x,y) & {\partial_{2} F}(x,y)} \right].
$$
If $ {\partial_{2} F}(a,b) \neq 0 $, then the Implicit Function Theorem implies that there exist

  
*
  
*an open interval $ U $ containing $ a $,
  
*an open interval $ V $ containing $ b $ and
  
*a continuously differentiable function $ f: U \to V $

such that

  
*
  
*$ f(a) = b $ and
  
*$ \{ (x,y) \in U \times V ~|~ F(x,y) = c \} = \{ (x,f(x)) \in \mathbb{R}^{2} ~|~ x \in U \} $.

Now, define a function $ G: U \to \mathbb{R} $ by
$$
\forall x \in U: \quad G(x) \stackrel{\text{def}}{=} F(x,f(x)).
$$
Clearly, we have
$$
\forall x \in U: \quad G(x) = c.
$$
It thus follows from the Multivariable Chain Rule that
\begin{align}
\forall x \in U: \quad
0 &= G'(x) \\
  &= {\partial_{1} F}(x,f(x)) \cdot 1 + {\partial_{2} F}(x,f(x)) \cdot f'(x) \\
  &= 0.
\end{align}
As $ {\partial_{2} F}(a,b) \neq 0 $, we therefore obtain
\begin{align}
f'(a) &= - \frac{{\partial_{1} F}(a,f(a))}{{\partial_{2} F}(a,f(a))} \\
      &= - \frac{{\partial_{1} F}(a,b)}{{\partial_{2} F}(a,b)}.
\end{align}
