# Consequence of Implicit function theorem?

Let $f: \mathbb{R}^n \rightarrow \mathbb{R}$ and assume that locally $\partial_1f(x)\ge C > 0$ for some constant $C>0.$

Then I saw an application of the implicit function theorem that claimed that this would imply locally

$f(x)=g(x)(x_1-h(x_2,...,x_n))$ for some functions $g,h$ however I do not see how this follows from the implicit function theorem. Can anybody shed some light on how this follows from the theorem?

To get an explicit representation of this special form you probably need to assume that $f(a)= 0$ for some $a$.
In this case the implicit function theorem tells you that, in a neighbourhood $U$ of $a$, the zero set $L_0:=\{x\in U:f(x)= 0\}$ can be written as
$$L_0 = \{(x_1,\ldots, x_n):x_1 - h(x_2,\ldots, x_n) = 0\}$$ You can then define $$g(x) = \frac{f(x)}{x_1 - h(x_2,\ldots, x_n)}$$ whenever $x\in U$ and $f(x)\neq 0$ which is equivalent to the fact that the denominator in that expression is not $=0$ (so it's well defined).
The point is that it does not matter how $g$ is defined on $L_0$, since $x_1 - h(x_2,\ldots, x_n)= 0$ on that set. This says nothing about regularity of $g$, though.
If there is no such $a$ then you can repeat the same with $f(x)-f(a)$ but with a slightly different result.