Implicit function theorem conclusion notation? I am working through implicit function theorem for the first time, and I have the following understanding. Given a system of $n$ equations,
\begin{equation}
f_i(x_1,\dots ,x_m,y_1,\dots , y_n)=0,\ \ \ \ \ \ \ i=1,\dots ,n
\end{equation}
With some point $p_0$ with coordinates $(a_1,\dots ,a_m,b_1,\dots ,b_n)$, and the condition,
\begin{equation}
\frac{\partial (f_1,\dots,f_n)}{\partial (y_1,\dots ,y_n)}\bigg|_{p_{0}}\neq 0
\end{equation}
Then the system of equations can be solved for $y_1,\dots ,y_n$ as functions of $x_1,\dots ,x_m$ in the neighbourhood of the point. In this case the following equations hold,
\begin{equation}
f_i(x_1,\dots ,x_m,y_1(x_1,\dots ,x_m),\dots ,y_n(x_1,\dots ,x_m))=0,\ \ \ \ \ i=1,\dots ,n
\end{equation}
However I do not understand the following notation, which is the conclusion of the theorem!
\begin{equation}
\frac{\partial y_i}{\partial x_j}\bigg|_{i\neq j}=-\frac{\frac{\partial(f_1,f_2,\dots ,f_n)}{\partial (y_1,\dots ,x_j,\dots ,y_n)}}{\frac{\partial (f_1,f_2,\dots ,f_n)}{\partial (y_1,\dots ,y_i,\dots ,y_n)}}
\end{equation}
Could you please explain what these Jacobian determinants are, and why the $x_j$ and $y_i$ are in the denominators of each expression? Many thanks! 
 A: I happen to have some notes on this, perhaps they help:
Given $n$-equations in $(m+n)$-unknowns when can we solve for the last $n$-variables as functions of the first $m$-variables ? Given a continuously differentiable mapping $G=(G_1,G_2,\dots , G_n): \mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ study the level set: (here $k_1,k_2,\dots , k_n$ are constants)
\begin{align} \notag
G_1(x_1, \dots , x_m, y_1, \dots , y_n)&=k_1 \\ \notag
G_2(x_1, \dots , x_m, y_1, \dots , y_n)&=k_2 \\ \notag
 & \vdots \\ \notag
G_n(x_1, \dots , x_m, y_1, \dots , y_n)&=k_n \notag
\end{align}
We wish to locally solve for $y_1, \dots , y_n$ as functions of $x_1, \dots x_m$. That is, find a mapping $h : \mathbb{R}^m \rightarrow \mathbb{R}^n$ such that $G(x,y)=k$ iff $y=h(x)$ near some point $(a,b) \in \mathbb{R}^m \times \mathbb{R}^n$ such that $G(a,b)=k$. In this section we use the notation $x=(x_1,x_2,\dots x_m)$ and $y=(y_1,y_2,\dots , y_n)$. 
Before we turn to the general problem let's analyze the unit-circle problem in this notation. We are given $G(x,y)=x^2+y^2$ and we wish to find $f(x)$ such that $y=f(x)$ solves $G(x,y)=1$. Differentiate with respect to $x$ and use the chain-rule:
$$ \frac{\partial G}{\partial x}\frac{dx}{dx} + \frac{\partial G}{\partial y}\frac{dy}{dx} = 0 $$
We find that $\boxed{dy/dx = -G_x/G_y} = -x/y$. Given this analysis we should suspect that if we are given some level curve $G(x,y)=k$ then we may be able to solve for $y$ as a function of $x$ near $p$ if $G(p)=k$ and  $G_y(p) \neq 0$. This suspicion is valid and it is one of the many consequences of the implicit function theorem.
We again turn to the linearization approximation. Suppose $G(x,y)=k$ where $x \in \mathbb{R}^m$ and $y \in \mathbb{R}^n$ and suppose $G: \mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}^n$ is continuously differentiable. Suppose $(a,b) \in \mathbb{R}^m \times \mathbb{R}^n$ has $G(a,b)=k$. Replace $G$ with its linearization based at $(a,b)$:
$$ G(x,y) \approx k + G'(a,b)(x-a,y-b) $$
here we have the matrix multiplication of the $n \times (m+n)$ matrix $G'(a,b)$ with the $(m+n) \times 1$ column vector $(x-a,y-b)$ to yield an $n$-component column vector. It is convenient to define partial derivatives with respect to a whole vector of variables,
$$ \frac{\partial G}{\partial x} = 
\left[ 
\begin{array}{ccc} 
 \tfrac{\partial G_1}{\partial x_1} & \cdots & \tfrac{\partial G_1}{\partial x_m}  \\ 
   \vdots &  & \vdots  \\
 \tfrac{\partial G_n}{\partial x_1} & \cdots & \tfrac{\partial G_n}{\partial x_m}
\end{array} 
\right] \qquad 
\frac{\partial G}{\partial y} = 
\left[ 
\begin{array}{ccc} 
 \tfrac{\partial G_1}{\partial y_1} & \cdots & \tfrac{\partial G_1}{\partial y_n}  \\ 
   \vdots &  & \vdots  \\
 \tfrac{\partial G_n}{\partial y_1} & \cdots & \tfrac{\partial G_n}{\partial y_n}
\end{array} 
\right] $$
In this notation we can write the $n \times (m+n)$ matrix $G'(a,b)$ as the concatenation of the $n \times m$ matrix $\frac{\partial G}{\partial x}(a,b) $ and the $n \times n$ matrix $\frac{\partial G}{\partial y}(a,b)$
$$ G'(a,b) = \biggl[\frac{\partial G}{\partial x}(a,b) \bigg{|} \frac{\partial G}{\partial y}(a,b) \biggl] $$
Therefore, for points close to $(a,b)$ we have:
$$ G(x,y) \approx k + \frac{\partial G}{\partial x}(a,b)(x-a)+\frac{\partial G}{\partial y}(a,b)(y-b) $$
The nonlinear problem $G(x,y)=k$ has been (locally) replaced by the linear problem of solving what follows for $y$:
$$ 
 k \approx k + \frac{\partial G}{\partial x}(a,b)(x-a)+\frac{\partial G}{\partial y}(a,b)(y-b) 
$$
Suppose the square matrix $\frac{\partial G}{\partial y}(a,b)$ is invertible at $(a,b)$ then we find the following approximation for the implicit solution of $G(x,y)=k$ for $y$ as a function of $x$:
$$ y = b - \biggl[\frac{\partial G}{\partial y}(a,b) \biggr]^{-1}\biggl[\frac{\partial G}{\partial x}(a,b)(x-a) \biggl]. $$
Of course this is not a formal proof, but it does suggest that $det\bigl[\frac{\partial G}{\partial y}(a,b) \bigr] \neq 0$ is a necessary condition for solving for the $y$ variables.  
As before suppose $G: \mathbb{R}^m \times \mathbb{R}^n \rightarrow \mathbb{R}^n$. Suppose we have a continuously differentiable function $h: \mathbb{R}^m \rightarrow \mathbb{R}^n$ such that $h(a)=b$ and $G(x,h(x))=k$. We seek to find the derivative of $h$ in terms of the derivative of $G$. This is a generalization of the implicit differentiation calculation we perform in calculus I. I'm including this to help you understand the notation a bit more before I state the implicit function theorem. Differentiate with respect to $x_l$ for $l \in \{1,2,\dots n\}$: 
$$  \frac{\partial}{\partial x_{l}} \biggl[ G(x,h(x)) \biggr] =  
\sum_{i=1}^{m}\frac{\partial G}{\partial x_i } \frac{\partial x_i}{\partial x_l } + 
\sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_l} = 
\frac{\partial G}{\partial x_l }  + 
\sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_l} = 0$$
we made use of the identity $\frac{\partial x_i}{\partial x_k } = \delta_{ik}$ to squash the sum of $i$ to the single nontrivial term and the zero on the r.h.s follows from the fact that $\frac{\partial}{\partial x_l} (k)=0$. Concatenate these derivatives from $k=1$ up to $k=m$:
$$ \biggl[ \frac{\partial G}{\partial x_1 }  + 
\sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_1} \bigg{|}
\frac{\partial G}{\partial x_2 }  + 
\sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_2} \bigg{|} \cdots 
\bigg{|}
\frac{\partial G}{\partial x_m }  + 
\sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_m} \biggr] =
[0|0| \cdots |0]  $$
Properties of matrix addition allow us to parse the expression above as follows:
$$ \biggl[ \frac{\partial G}{\partial x_1 } \bigg{|}
\frac{\partial G}{\partial x_2 }  
\bigg{|} \cdots 
\bigg{|}
\frac{\partial G}{\partial x_m }  
 \biggr] + 
 \biggl[  
\sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_1} \bigg{|}
\sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_2} \bigg{|} \cdots 
\bigg{|}
\sum_{j=1}^{n}\frac{\partial G}{\partial y_j}\frac{\partial h_j}{\partial x_m} \biggr] 
=
[0|0| \cdots |0]  $$
But, this reduces to 
$$ \frac{\partial G}{\partial x } + \biggl[  
\frac{\partial G}{\partial y}\frac{\partial h}{\partial x_1} \bigg{|}
\frac{\partial G}{\partial y}\frac{\partial h}{\partial x_2} \bigg{|} \cdots 
\bigg{|}
\frac{\partial G}{\partial y}\frac{\partial h}{\partial x_m} \biggr]  = 0 \in \mathbb{R}^{m \times n} $$
The concatenation property of matrix multiplication states $[Ab_1|Ab_2| \cdots | Ab_m] = A[b_1|b_2| \cdots | b_m]$ we use this to write the expression once more,
$$ \frac{\partial G}{\partial x } 
+ \frac{\partial G}{\partial y} \biggl[  
\frac{\partial h}{\partial x_1} \bigg{|}
\frac{\partial h}{\partial x_2} \bigg{|} \cdots 
\bigg{|}
\frac{\partial h}{\partial x_m} \biggr]  = 0  \ \ \Rightarrow \ \ \frac{\partial G}{\partial x } 
+ \frac{\partial G}{\partial y} 
\frac{\partial h}{\partial x}  = 0 \ \ \Rightarrow \ \ \boxed{\frac{\partial h}{\partial x} = -\frac{\partial G}{\partial y}^{-1}\frac{\partial G}{\partial x }} $$
where in the last implication we made use of the assumption that $\frac{\partial G}{\partial y}$ is invertible. 
