Proving $\nabla(f,(xy))$ is a linear isomorphism 
Let $f:\mathbb{R}^2\to\mathbb{R}^2,(x,y)\to(x^2-y^2,2xy)$. Prove that $f$ is not injective but $f$ is a local difeomorphism in all the points of $\mathbb{R}^2$

I decided to use the Inverse function theorem, stated:

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.

My strategy is to prove that $\nabla(f,(x,y))$ is a linear isomorphism such that I can apply the Inverse function theorem to conclude that $f$ is an open map and a difeomorphism.
I computed the gradient:
$\nabla(f,(xy))=\begin{bmatrix}2x&-2y\\2y&2x\end{bmatrix}$
$\det\nabla(f,(xy))=4x^2+4y^2>0$ 
I know what is a isomorphism from linear algebra(a bijective linear function between two vector spaces). 
Question:
On this case I am not seeing how to prove $\nabla(f,(xy))$ is a linear isomorphism. How can I do it? Can someone provide me a proof? 
Thanks in advance!
 A: The map $f:\mathbb{R^2}\rightarrow \mathbb{R}^2$ $(x,y)\mapsto (x^2-y^2, 2xy)$ is indeed not injective, and hence not a global diffeomorphism. 
Consider the points $(x,y)=(-1,1), (-1,1)$. Both points map to $(0,-2)$ in the range. We can actually classify all points where injectivity fails. 
In order to have a diffeomorphism, we start by considering the Jacobian $\nabla(f)=\begin{bmatrix}2x&-2y\\2y&2x\end{bmatrix}$. Note, the Jacobian is invertible when $det(\nabla(f))\neq 0$, so when $4x^2-4y^2\neq 0$, Which reduces to $x^2=y^2$. 
According to the Inverse Function Theorem, the map $f$ is a diffeomorphism on the set $S\subset \mathbb{R}^2$, $S=\{(x,y)\in \mathbb{R}^2: x^2\neq y^2$}. 
Note that the map $f$ is a diffeomorphism on $S$ since when $f$ is restricted to $S$, we have an injective map. 
EDIT
To show surjectivity, consider that we consider the map $(x^2-y^2,2xy)=(a,b)$ and we want this to be well-defined $\forall a,b\in \mathbb{R}$. We can "solve component-wise". We can set $x^2-y^2=a$, and note for any $a$, for any $x$ (WLOG), we have $x^2=y^2-a$ and $x=\sqrt{y^2-a}$, so any $y$ for which this expression is defined will do (square root is non-negative). So if we restrict ourselves to $x\in \mathbb{R}^{+}$, and where the square root is non-negative, namely $y^2\geq a$, we get surjectivity. 
for $2xy=b$ is easier, once we solve for $x,y$ for the first component, We can now just consider $k=xy$ a fixed constant, and $2k=b$ is well-defined. 
Taking into account on the domain restrictions to obtain injectivity, and surjectivity, we have the domain where we have a linear isomorphism/diffeomorphism. 
