Proving $(x,y,z)\to \dfrac{(x,y,z)}{x^2+y^2+z^2}$ is diffeomorphism of $\mathscr{C}^{\infty}$ class 
Let $K:\mathbb{R}^3 \setminus \{0\} \to \mathbb{R}^3 \setminus \{0 \}$ be defined by,
$$(x,y,z)\mapsto \frac{1}{x^2+y^2+z^2}(x,y,z)$$
Show that $K$ is a diffeomorphism of class$\mathscr{C}^{\infty}$.

My strategy is to prove that $J((f,(x,y))$, the jacobian matrix of $f$ at $(x,y)$, is a linear isomorphism such that I can apply the Inverse function theorem to conclude that $f$ is an open map and a difeomorphism.
Inverse function theorem:

Theorem: Let $\Omega$ be an open set of $\mathbb{R}^{n}$ and let $f:\Omega\to\mathbb{R}^n$ be a function of $\mathscr{C}^r(r\in\mathbb{N}\cup\{\infty\})$. Consider $a\in\Omega$ such as $f´(a):\mathbb{R}^n\to\mathbb{R}^n$ is a linear isomorphism and $b=f(a)$. Therefore there exists an open neighbourhood $U$ of $a$ in $\Omega$ and there is an open neighbourhood $V$ of $b$ in $\mathbb{R}^n$ such that $f(U)=V$ and $f:U\to V$ is a difeomorphism of $\mathscr{C}^r$ class.

$J(f,(x,y,z))=\begin{bmatrix}-\frac{2x}{x^2+y^2+z^2}\\-\frac{2y}{x^2+y^2+z^2}\\-\frac{2z}{x^2+y^2+z^2}\end{bmatrix}$
I need to prove $\det J(f,(x,y,z))\neq 0$.
Question:
1) How can I compute $\det J(f,(x,y,z))\neq 0$? Is there a way to prove the different lines of the matrix are linearly independent?
2)
I understand that a function of $\mathscr{C}^{\infty}$ is function infinitely differentiable. However if I differentiate $f(x,y,z)$ two times more I get the null matrix. What is $\mathscr{C}^{\infty}$ class about?
Thanks in advance!
 A: It's just as easy to do this in any $\mathbb R^n,$ where we write points as $x=(x_1,\dots , x_n).$ Define $U=\mathbb R^n\setminus \{0\}.$ We want to consider the map $F: U\to U$ defined by $F(x) = x/|x|^2.$
I'm not sure why you are having trouble with the $C^\infty$ issue. The components of $F$ have the form $x_k/|x|^2.$ This is a quotient of polynomials, with the denominator never vanishing on $U.$ Such a function belongs to $C^\infty(U),$ hence so does $F.$
Let's now observe $F\circ F(x) = x$ on $U,$ an easy computation. Immediately we see $F$ is a bijection of $U$ onto $U.$
The last observation allows us to skip tedious computations in showing $DF(x)$ has the desired properties. For any $x\in U,$ the chain rule gives
$$\tag 1 D (F\circ F)(x) = DF(F(x)) \circ DF(x).$$
Since $F\circ F = I,$ the identity, and $DI = I,$  the left side of $(1)$ equals $I.$ Thus the kernel of $DF(x)$ is $\{0\},$ showing $DF(x)$ is nonsingular. (Therefore $\text { det} \,J(F,x) \ne 0,$ although we don't need this fact.)
Summarizing, we have shown $F:U\to U$ is a $C^\infty$ bijection, with $DF(x)$ nonsingular everywhere in $U.$ Since $F^{-1} = F,$ the same is true for the inverse map, hence $F$ is a $C^\infty$ diffeomorphism of $U$ onto $U.$
