I believe this will be a very long answer if anyone tries to write the full proof or anything so I'll specify which specific parts I am having trouble with to save people's time.
I want prove three things:
First we know, $f: U \to V$ where $U \subset \mathbb{R^n}$ and $V \subset \mathbb{R^n}$. And let $a \in U$, s.t. $Df(a)$ is invertible.
(1) $f$ is one-to-one and onto (hence invertible)
(2) The inverse function $f^{-1}$ is of class $C^1$
(3) If $x \in \mathbb{R^n}$ and $y=f(x)\in\mathbb{R^n}$ then $D(f^{-1})(y)=(Df(x))^{-1}$.
Proof: So I know that the (2) follows directly from one of the conclusions in the implicit function theorem.
I am needing help in proving that $f$ is invertible. I know from implicit function theorem that there is a function $g$ such that $g(f(a))=a$. My idea was to prove that $f(g(b))=b$? But I'm not sure if this will work and if this approach can work not sure how to make it work. If this works we know that $g=f^{-1}$.
Also, for (3) what I got to was $Dg(y)=Dg(f(x))Df(x)$ by chain rule. And since we know that $g=f^{-1}$ we can simplify the above equation to: $Dg(y)=DxDf(x)$ but not sure how to work with the $DxDf(x)$ to $(Df(x))^{-1}$.
Explicit hints will be greatly appreciate, thanks in advance.