Inversion map is a Conformal map

Question

I'm studying PDE by Evans book and I need to show that the inversion map $f:\mathbb{R}^n-\{0\}\to \mathbb{R}^n$, defined by

$$f(x)=\frac{x}{\|x\|^2}$$

is conformal. So I have a hint, show that $Df.(Df)^{t}=\|x\|^{-4}I.$

Fisrt: I don't understood why that hint solved my problem, cause I know that $g:\mathbb{R}^n\to\mathbb{R}^n$ is conformal if for $x,y\in\mathbb{R}^n$ we have

$$Dg(x).(Dg(y))^t=C(x) x.y^t$$ ($x$ is a colunn vector) and $C(x)>0$ positive function, all right or not?!

Second: Anyway, I used of hint and I found

$$Df.(Df)^t=\|x\|^{-4}\left(I-\frac{4}{\|x^2\|}x.x^t+\frac{4}{\|x\|^4}x.x^t.x.x^t \right)$$ but I don't know how to conclude the affirmation. Can someone help me? Thanks!

Answer 1

First: The definition of conformality looks odd; did you mean something like $Dg(x)^{t} \cdot Dg(x) = C(x) I$, i.e., $$ \left\langle Dg(x)u, Dg(x)v \right\rangle = C(x) \langle u, v\rangle \quad\text{for all $u$, $v$} $$ instead?

Second: Inside the parentheses, note that $x^{t}x = \|x\|^{2}$.