In Analysis 2 by Terence Tao, the Implicit function theorem is stated as follow:
Let $E$ be an open subset of $\mathbb{R^n}$. Let $f: E \to \mathbb{R}$ be continuously differentiable.
Let $y = (y_1, y_2, .., y_n)$ be a point in $E$ such that $f(y) = 0$ and $\frac{\partial f}{\partial x_n} (y) \neq 0$.
Then:
$\color{red}{(1)}$. There exists an open subset $U$ of $\mathbb{R^{n-1}}$ containing the point $(y_1,..,y_{n-1})$ and an open subset $V$ of $E$ containing the point $y$ and a function $g: U \to \mathbb{R}$ such that:
+) $g(y_1,..,y_{n-1}) = y_n \qquad$ and
+) The set {$(x_1, x_2,..,x_n) \in V$ : $f(x_1,...,x_n) = 0 $} = {$(x_1,..,x_{n-1}, g(x_1,..,x_{n-1})$ : $(x_1,x_2,..,x_{n-1}) \in U$ }
$\color{red}{(2)}$. Function $g$ is differentiable at the point $(y_1, y_2, ..., y_{n-1})$
$\color{red}{(3)}$. We have $\frac{\partial g}{\partial x_j} (y_1,..,y_{n-1})$ = $- \frac{\partial f}{\partial x_j}(y) / \frac{\partial f}{\partial x_n}(y)$ $\qquad$ for all $1 <= j <= n-1$
I have understood the author's proof of the point $(1)$ and $(2)$ but don't know how to use the chain rule (which relates to total derivative), differentiating in $x_j$ (as suggested by the author) to prove (3).
Theorem 6.4.1 (Several variable calculus chain rule)
Let $E$ be a subset of $\mathbb{R^n}$, and let $F$ be a subset of $\mathbb{R^m}$. Let $f : E \to F$ be a function, and let $g : F \to \mathbb{R^p}$ be another function. Let $x_0$ be a point in the interior of $E$.
Suppose that $f$ is differentiable at $x_0$, and that $f(x_0)$ is in the interior of $F$ . Suppose also that $g$ is differentiable at $f(x_0)$.
Then $g ◦ f : E \to \mathbb{R^p}$ is also differentiable at $x_0$, and we have the formula:
$(g◦f)'(x_0) = g'(f(x_0)) f'(x_0)$
Could you please give a sketch of a proof of point (3) using the chain rule ?
EDIT: As stated by the author, what we want is to have the equation:
$\frac{\partial f}{\partial x_j}(y)$ + $\frac{\partial f}{\partial x_n}(y). \frac{\partial g}{\partial x_j} (y_1,..,y_{n-1}) $ $= 0$