In Implicit Function Theorem, we can find a continously differentiable function $g$ which satisfies $f(x,g(x))=0 \quad(1),$ also we have $[Dg] = -[D_{1}f \cdots D_{k}f]^{-1}[D_{k+1}f \cdots D_{n}f] \quad(2),$ while $1,\cdots,k$ corresponds to pivots columns while $k+1,\cdots, n$ corresponds to non-pivot one.
My question is how can we prove $(2)?$
I plan to use the chain rule on $f,$ i.e. $$ \dfrac{\partial f(x,g(x))}{\partial x_{j}} = \frac{\partial f(x,g(x))}{\partial y} \times \frac{\partial g}{\partial x_{j}} $$ But I don't know how to proceed after that.
I think it is good to differentiate with respect to $x_j$'s from the equation :
$$ f(x,g(x))= 0.$$
Let :$$\mathbf{y} = \begin{bmatrix} y_1 \\ \vdots \\ y_m \end{bmatrix} = \begin{bmatrix} g_1(x_1,...,x_n)\\ \vdots \\ g_m(x_1,....,x_n)\end{bmatrix}.$$
Let $f = \begin{bmatrix}f_1 \\ \vdots \\ f_m\end{bmatrix}$. Now if we differentiate $f_i$ with respect to $x_j$, when $f(x,g(x))=0$, we will have :
$$ \frac{\partial f_i}{\partial x_j} + \sum_{k=1}^{m} \frac{\partial g_k}{\partial x_j}\times \frac{\partial f_i}{\partial y_k} = 0. $$
So fixing $j$ and varying $i$. you will get :
$$ \begin{bmatrix}-\frac{\partial f_1}{\partial x_j} \\ \vdots \\ -\frac{\partial f_m}{\partial x_j}\end{bmatrix} = \begin{bmatrix} \sum_{k=1}^{m} \frac{\partial f_1}{\partial y_k} \times \frac{\partial g_k}{\partial x_j}\\ \vdots \\ \sum_{k=1}^{m} \frac{\partial f_m}{\partial y_k} \times \frac{\partial g_k}{\partial x_j} \end{bmatrix}. \\ \begin{bmatrix}-\frac{\partial f_1}{\partial x_j} \\ \vdots \\ -\frac{\partial f_m}{\partial x_j}\end{bmatrix} = \begin{bmatrix}\frac{\partial f_1}{\partial y_1} & \dots & \frac{\partial f_1}{\partial y_m} \\ \vdots & \dots & \vdots \\ \frac{\partial f_m}{\partial y_1} & \dots & \frac{\partial f_m}{\partial y_m} \end{bmatrix} \times \begin{bmatrix} \frac{\partial g_1}{\partial x_j} \\ \vdots \\ \frac{\partial g_m}{\partial x_j} \end{bmatrix} $$
Now since $D_{\mathbb y} f$ is invertible at the points of the equation $f(x,g(x))=0$, we get our desired result.
Note : We are doing this differentiation at some point $(\bf a , \bf b)$ with $f(\bf a , \bf b) = 0.$ and that $g$ is defined on an open set $U$ around $\bf a$.