Derivatives of component inverse functions I might have missed the point of the following questions. Anyone kindly give a suggestion?

Let $f:\mathbb{R}_\mathbf{x}^3\to\mathbb{R}_\mathbf{y}^3$ and
  $g:\mathbb{R}_\mathbf{y}^3\to\mathbb{R}_\mathbf{x}^3$ be
  $\mathscr{C}^1$ inverse functions. Show that
$$ \frac{\partial g_1}{\partial y_1}=\frac{1}{J} \frac{\partial (f_2 ,
 f_3)}{\partial (x_2, x_3)},J=\frac{\partial (f_1 , f_2 ,
 f_3)}{\partial (x_1 , x_2 , x_3)} $$
And obtain similar formulas for the other derivatives of the component
  functions of $g$.

A similar question in terms of implicit functions:

If the equations $f(x,y,z)=0$, $g(x,y,z)=0$ can be solved for $y$ and
  $z$ as differentiable functions of $x$, show that  \begin{eqnarray*}
 \frac{dy}{dx}=\frac{1}{J}\frac{\partial(f,g)}{\partial(z,x)}, &  &
 \frac{dz}{dx}=\frac{1}{J}\frac{\partial(f,g)}{\partial(x,y)},
 \end{eqnarray*} where $J=\frac{\partial(f,g)}{\partial(y,z)}$.

which I have just solved as below.
 A: In general, if $f\circ g=\cal I$, where $\cal I:\mathbb R^n\to\mathbb R^n$, is the identity, then, accoriding to the Chain Rule for vector valued functions
$$
\cal I=D\cal I=D(f\circ g)(x) =Df\big(g(x)\big)\cdot Dg(x),
$$
where $Dh$ is the differential of $h$, and hence
$$
1=\det \cal I=\det Df\big(g(x)\big)\cdot \det Dg(x).
$$
Thus
$$
\det Dg(x)=\frac{1}{\det Df\big(g(x)\big)}.
$$
A: I myself tried the second question just now.
Define $f(x,y(x),z(x))=0$ and $g(x,y(x),z(x))=0$. Then by the chain rule, 
\begin{eqnarray*}
\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial z}\frac{dz}{dx}=0 & \implies & \frac{\partial f}{\partial x}\frac{\partial g}{\partial z}+\frac{\partial f}{\partial y}\frac{\partial g}{\partial z}\frac{dy}{dx}+\frac{\partial f}{\partial z}\frac{\partial g}{\partial z}\frac{dz}{dx}=0\\
\frac{\partial g}{\partial x}+\frac{\partial g}{\partial y}\frac{dy}{dx}+\frac{\partial g}{\partial z}\frac{dz}{dx}=0 & \implies & \frac{\partial f}{\partial z}\frac{\partial g}{\partial x}+\frac{\partial f}{\partial z}\frac{\partial g}{\partial y}\frac{dy}{dx}+\frac{\partial f}{\partial z}\frac{\partial g}{\partial z}\frac{dz}{dx}=0
\end{eqnarray*}
So 
\begin{eqnarray*}
 &  & -\left(\frac{\partial f}{\partial z}\frac{\partial g}{\partial x}-\frac{\partial f}{\partial x}\frac{\partial g}{\partial z}\right)+\left(\frac{\partial f}{\partial y}\frac{\partial g}{\partial z}-\frac{\partial f}{\partial z}\frac{\partial g}{\partial y}\right)\frac{dy}{dx}=0\\
 & \implies & -\frac{\partial(f,g)}{\partial(z,x)}+\frac{\partial(f,g)}{\partial(y,z)}\frac{dy}{dx}=0\\
 & \implies & \frac{dy}{dx}=\frac{1}{\frac{\partial(f,g)}{\partial(y,z)}}\frac{\partial(f,g)}{\partial(z,x)}=\frac{1}{J}\frac{\partial(f,g)}{\partial(z,x)}
\end{eqnarray*}
It can be solved similarly for $\frac{dz}{dx}$.
