I want to show the following:
Let $G \subset \mathbb{R}^n$ be open, $f \in C^1(G)$ and $\det J_f(x) \neq 0$ for all $x \in G$. Then $f$ is an open map, i.e. the image $f(O)$ of each open subset $O$ of $G$ is open.
The following hint is given:
Let $O \subset G$ be open. Apply at the function $F : O\times \mathbb{R}^n \rightarrow \mathbb{R}^n$ with $F(x,y) = f(x)-y$ and $f(x_0) = y_0$; $x_0 \in O$ the proposition of implicit function.
$$$$
Using the function of the hint the condition of the proposition are satisfied, aren't they?
Do we not get from that sentence that we can uniquely solve for $y_0$ the relation $F(x_0,y_0)=0$, or not?
But what do we get from that?
It is much more natural to solve this problem using the inverse function theorem. If $x\in G$, then, since $f'(x)$ is invertible, that theorem assures us that there are open neighbourhoods $U$ and $V$ of $x$ and of $f(x)$ respectively such that the restriction of $f$ to $U$ is a diffeomorphism from $U$ onto $V$. In particular, $f(U)=V$. So, if $O$ is an open subset of $G$, $f(O)$ is a neighbourhood of each of its points and therefore it's an open set.