Show that the given function is a diffeomorphism 
Let $U=\{x\in\mathbb{R}^n: ||x||<1\}$. If we define $f:U\rightarrow\mathbb{R}^n$ by $f(x) = \displaystyle\frac{x}{\sqrt{1-||x||^2}}$, show that $f$ is a diffeomorphism and $f^{-1}:\mathbb{R}^n\rightarrow U$ is given by $f(y) = \displaystyle\frac{y}{\sqrt{1-||y||^2}}$.

(1) Proving bijectiviy
(1.1) Injectivity
$f(x) = f(y) \iff \displaystyle\frac{x_i}{\sqrt{1-||x||^2}}=\displaystyle\frac{y_i}{\sqrt{1-||y||^2}} \iff x_i^2[1-(y_1^2+\dots+y_n^2)] =y_i[1-(x_1^2+\dots+x_n^2)]$.
Now adding all of the following identities
$\begin{cases} x_1^2 -x_1^2y_1^2+\dots+x_1^2y_n^2 = y_1^2-x_1^2y_1^2+\dots+x_n^2y_1^2 & (1)\\ \dots\\ x_n^2 -x_n^2y_1^2+\dots+x_n^2y_n^2 = y_n^2-x_1^2y_n^2+\dots+x_n^2y_n^2 & (n)\end{cases}$
I have $x_1^2+\dots+x_n^2 = y_1^2+\dots+y_n^2 \implies ||x||=||y||$. Then $\displaystyle\frac{x_i}{\sqrt{1-||x||^2}}=\displaystyle\frac{y_i}{\sqrt{1-||y||^2}} \iff \displaystyle\frac{x_i}{\sqrt{1-\alpha}}=\displaystyle\frac{y_i}{\sqrt{1-\alpha}}\iff x_i=y_i\implies x=y$.
(1.2) Surjectiviy
I'm not sure how should I do this, see (3.3)
(2) Proving that $f$ is differentiable applying the quotient rule to the coordinate functions
If $i\neq j$ then $\displaystyle\frac{\partial f_i}{\partial x_j}= \displaystyle\frac{\partial}{\partial x_j}\left( \displaystyle\frac{x_i}{1-||x||^2}\right) = \displaystyle\frac{\frac{2x_i}{2\sqrt{1-||x||}}}{{1-||x||}} = \displaystyle\frac{x_i}{(1-||x||^2)^{3/2}}$ which is continous for every $x\in U$.
And $\displaystyle\frac{\partial f_i}{\partial x_i} = \displaystyle\frac{\partial}{\partial x_i}\left( \displaystyle\frac{x_i}{1-||x||^2}\right)\displaystyle\frac{\sqrt{1-||x||^2}-2x_i^2}{1-||x||^2} = \displaystyle\frac{1}{\sqrt{1-||x||^2}} - \displaystyle\frac{2x_i^2}{1-||x||^2}$ which is also continuous if $x\in U$, right?.
By the continuity of the partial derivatives for $f_i$ I conclude that $f_i$ is differentiable and since it applies for every $i=1,\dots,n$ I conclude $f$ that is differentiable.
(3) Some questions.
I'm not sure how to proceed from here, I have a few guesses and questions:
(3.1) The differential for $f$ will be a matrix with values computed above, right?. I can use the inverse function theorem to find the inverse of $f$ by $(f^{-1})'(y)=1/(f'(y))^{-1}$, but how does it works? -I mean, am I supposed to divide by the matrix?-.
(3.2) Even if I get $f^{-1})'$, how can I get $f^{-1}$. Should I integrate the coordinate functions?
(3.3) I'm not sure how to prove the surjectivity either, could I use that $\operatorname{dim}(U)=\operatorname{dim}(\mathbb{R}^n)$ the injectivity implies surjectivity?. But wouldn't I need a basis for $U$?, can I use the canonical basis with the restriction that $\sum_i x_i^2 < 1$?.
 A: If we define
$$
g:\mathbb{R}^n \to U,\ g(x)=\frac{x}{\sqrt{1+\|x\|^2}},
$$
then for every $y \in \mathbb{R}^n$ we have
\begin{eqnarray}
f(g(y))&=&\frac{g(y)}{\sqrt{1-\|g(y)\|^2}}=\left(1-\frac{\|y\|^2}{1+\|y\|^2}\right)^{-1/2}\cdot\frac{y}{\sqrt{1+\|y\|^2}}\\
&=&\sqrt{1+\|y\|^2}\cdot\frac{y}{\sqrt{1+\|y\|^2}}=y.
\end{eqnarray}
Similarly, for every $x \in U$ we have
\begin{eqnarray}
g(f(x))&=&\frac{f(x)}{\sqrt{1+\|f(x)\|^2}}=\left(1+\frac{\|x\|^2}{1-\|x\|^2}\right)^{-1/2}\cdot\frac{x}{\sqrt{1-\|x\|^2}}\\
&=&\sqrt{1-\|x\|^2}\cdot\frac{x}{\sqrt{1-\|x\|^2}}=x.
\end{eqnarray}
Hence $f$ is bijective, and $f^{-1}=g$.
You can prove that $f$ and $g$ are differentiable, therefore $f$ is a diffeomorphism.
A: Sometimes, it is useful to keep vector notation rather than writing $f = (f_1,...,f_n)$.
Easier way to prove injectivity: Let be $a_w = \sqrt{1-||w||^2}$.
$$\frac{x}{\sqrt{1-||x||^2}} = \frac{y}{\sqrt{1-||y||^2}}\Leftrightarrow a_yx = a_x y\text{,}$$
therefore $x$ and $y$ are colinear. If $x=\vec{0}$, then it follows that $y =\vec{0}$, because $a_x, a_y\neq0$. So let $x$ be nonzero vector. It follows that $y$ is nonzero vector and $y = tx$ for a $t = \frac{a_y}{a_x}>0$.
Function $g: [0,1) \to \mathbb{R},\; g(u^2) = \frac{1}{\sqrt{1-u^2}}$ is injection, therefore $||x|| =||y ||$, so $x=y$, since we already know that they are colinear and $t>0$.
Surjectivity:
Choose arbitrary $z\in \mathbb{R}^n$. You have to find $x$ with the right direction and the right norm. If $z=\vec{0}$, then $f(\vec{0})=z$. Otherwise
$$z = \frac{x}{a_x} \Rightarrow ||z||\sqrt{1-||x||^2} = ||x||\text{.}$$
From here, you should express the norm $T = ||x||$ and prove that $T<1$. It follows that
$$x = T \frac{z}{||z||}$$
is the desired vector.
Smoothness:
identity on $\mathbb{R}^n$ is smooth. Function $$g_1(u) = 1-u$$ is smooth. Function $g_2:(0,\infty)\to \mathbb{R}$, $$ \; g_2(u) = \frac{1}{\sqrt{u}}$$ is also smooth. Therefore $g$ is smooth since $g = g_2 \circ g_1$. It is also known that $x\to ||x||^2$ is smooth function (and you can easily prove it). Therefore $f(x) = g(||x^2||)x$ is smooth (again, you can easily show that function $(t,\vec{x})\to t\vec{x}$ is smooth).
Dividing by matrix: you cannot devide by matrix, but you can multiply by its inverse. And it holds the analogue of the theorem you are mentioning in 3.1.
