Product of Jacobian and First Partial of Inverse Function We are given the question of showing the following:
\begin{equation}
J_{\textbf{f}}(\vec{x})D_1g_i(\vec{y})=
\begin{vmatrix}
\delta_{i,1} & D_1f_2(\vec{x}) & D_1f_3(\vec{x}) \\
\delta_{i,2} & D_2f_2(\vec{x}) & D_2f_3(\vec{x}) \\
\delta_{i,3} & D_3f_2(\vec{x}) & D_3f_3(\vec{x})
\end{vmatrix}
\end{equation}
Where $\textbf{g}$ is the inverse of $\textbf{f}$, and $\delta_{i,j}=1$ if $i=j$ or $0$ otherwise. $\vec{y}=\textbf{f}(\vec{x})$, so $\textbf{g}$ is the inverse of $\textbf{f}$.
I've found $D_1g_1$ as $D_1g_1D_1f_1+D_2g_1D_1f_2+D_3g_1D_1f_3$, based on the chain rule.
I've also found the Jacobian Determinant as 
\begin{equation}
J_{\textbf{f}}(\vec{x})=\begin{vmatrix}
D_1f_1(x) & D_1f_2(x) & D_1f_3(x) \\
D_2f_1(x) & D_2f_2(x) & D_2f_3(x) \\
D_3f_1(x) & D_3f_2(x) & D_3f_3(x)
\end{vmatrix}
\end{equation}
So, if the statement is true, then in the $i=1$ case, $J_{\textbf{f}}D_1g_1(\textbf{f}(\vec{x}))=D_2f_2(x)D_3f_3(x)-D_2f_3(x)D_3f_2(x)$.
However, I'm not sure how to go about showing this. Once I calculate the derivative of $g_1$ and the Jacobian determinant of $\textbf{f}$, multiplying them gives a 27 term expression, and since they all have different factors of $g_i$, it doesn't appear that they will cancel down to what we want.
How can I show that the product of the Jacobian of $\textbf{f}$ and the first derivative of $g_1$ equals the given expression?
 A: $$\newcommand{\partialD}[2]{\frac{\partial #1}{\partial #2}}
\newcommand{\bvec}[1]{\textbf{#1}}$$
This was my attempt at an answer for the assignment. It definitely needs improvement, but maybe it can spark some discussion on this question and help me move towards something more correct.
From Apos. theorem 13.6, we have that $\textbf{D}\textbf{f}(x)\textbf{D}\textbf{g}(y)=\text{Id}$. That means then that a linear combination of $\textbf{D}f$ and the columns vectors of $\textbf{D}g$. 
Notation: Let $D_ig$ be the $i$-th column vector of $\textbf{D}g$. Then, $D_1g_i$ is the first component of the first column vector.
Note that for any column vector $D_ig$, we have
$$\textbf{D}f(x)D_ig=\textbf{e}_i$$
Then, we have by Cramer's Rule
    \begin{equation}\label{eq1}
D_1g_i=\frac{\text{Det}\left[\textbf{D}f(x)\right]_i}{\text{Det}[\textbf{D}f]}
\end{equation}
Note that 
   $$\textbf{D}f(x)=\begin{bmatrix}
 D_1f_1(x) & D_2f_1(x) & D_3f_1(x)  \\
 D_1f_2(x) & D_2f_2(x) & D_3f_2(x)  \\
 D_1f_3(x) & D_2f_3(x) & D_3f_3(x)  \\
   \end{bmatrix}$$
Since the determinant of a matrix is equal to the dete of its transpose we can say, for example using $i=1$
$$
\text{Det}[\textbf{D}f]_1=
\begin{vmatrix}
    1 & D_2f_1(x) & D_3f_1(x)  \\
    0 & D_2f_2(x) & D_3f_2(x)  \\
    0 & D_2f_3(x) & D_3f_3(x)  \\ 
\end{vmatrix}=
\begin{vmatrix}
    1 & D_1f_2(x) & D_1f_3(x)  \\
    0 & D_2f_2(x) & D_2f_3(x)  \\
    0 & D_3f_2(x) & D_3f_3(x)  \\
\end{vmatrix}
=
\text{Det}([\textbf{D}f]^T)_i
   $$
Because $g(y)=g(f(x))=x$, we have that $D_1g_i=\partialD{g_i}{x_1}$. If $i=1$, then this is simply $\partialD{g_1}{x_1}=1$. If $i\not=1$, we get $0$, so in general $D_jg_i=\delta_{i,j}$.
(This is just hand-waving :() Furthermore, since any row exchanges applied to $\textbf{D}f$ in the numerator will negate any sign changes in the denominator, we may always transform the matrix to get the $e_i$ column into the first column.
Then \ref{eq1} becomes
   \begin{equation*}\label{eq2}
   D_1g_i=\frac{\begin{vmatrix}
\delta_{1,i} & D_1f_2(x) & D_1f_3(x)  \\
\delta_{2,i} & D_2f_2(x) & D_2f_3(x)  \\
\delta_{3,i} & D_3f_2(x) & D_3f_3(x)  \\
   \end{vmatrix}}{J_{\bvec{f}(x)}}
   \implies
\end{equation*}
\begin{equation}
 J_{\bvec{f}}(x)D_1g_i=\begin{vmatrix}
\delta_{1,i} & D_1f_2(x) & D_3f_1(x)  \\
\delta_{2,i} & D_1f_2(x) & D_3f_2(x)  \\
\delta_{3,i} & D_1f_3(x) & D_3f_1(x) 
   \end{vmatrix}
   \end{equation}
Then, for $i=1$
\begin{equation}
 J_{\bvec{f}}(x)D_1g_1=\begin{vmatrix}
1 & D_1f_2(x) & D_3f_1(x)  \\
0 & D_1f_2(x) & D_3f_2(x)  \\
0 & D_1f_3(x) & D_3f_1(x) 
   \end{vmatrix} = \frac{\partial(f_2,f_3)}{\partial(x_2,x_3)}
\end{equation}
and so 
\begin{equation}
D_1g_1=\begin{vmatrix}
1 & D_1f_2(x) & D_3f_1(x)  \\
0 & D_1f_2(x) & D_3f_2(x)  \\
0 & D_1f_3(x) & D_3f_1(x) 
   \end{vmatrix} = \frac{\frac{\partial(f_2,f_3)}{\partial(x_2,x_3)}}{J_{\bvec{f}}}=\frac{\frac{\partial(f_2,f_3)}{\partial(x_2,x_3)}}{\frac{\partial (f_1,f_2,f_3}{\partial (x_1,x_2,x_3)}}
\end{equation}
