$f:\mathbb{R}^2\to \mathbb{R}^2, (x,y)\mapsto \begin{pmatrix} x^3-xy^2\\x^2y-y^3 \end{pmatrix}$ Determine all points $ a \in \mathbb {R}^{2} $ for which open sets $ U, V \subset \mathbb {R}^{2} $ exist such that $f|_{U}: U \rightarrow V $ is a diffeomorphism. In this case, also determine $ D(f_{U}^{-1})(f(a) $.
What I use:
The Inverse Function Theorem $(*)$ $ f: \mathbb{R}^{n} \to \mathbb{R}^{n} $ be continuously differentiable. If $\det Df(a) \neq 0 $ then there exists an open set $ U $ with $a\in U$ and an open set $V$ with $f(a)\in V$ such that $ f: U \rightarrow V $ is a local $C^1$-diffeomorphism.
So there exists a inverse $ f^{-1}: V \to U $ which is differentiable for all $y \in V $.
Then $Df^{-1}(y)=[Df(f^{-1}(y))]^{-1} $.
Set $a=f^{-1}(y)$ then $Df^{-1}(y)=[Df(a)]^{-1} $
Progress:
$f$ is continuously partially differentiable, so $f$ is differentiable.
$Df((x,y))=\begin{pmatrix} 3x^2-y^2&-2xy\\2xy&x^2-3y^2 \end{pmatrix}\Rightarrow$ $Df$ is continuous, so $f$ is continuously differentiable.
$\det \begin{pmatrix} 3x^2-y^2&-2xy\\2xy&x^2-3y^2 \end{pmatrix}=\det Df((x,y))=3x^4-6x^2y^2+3y^4=3(x^4+y^4)-6x^2y^2=0\Rightarrow 3(x^4+y^4)=6x^2y^2$
Set $x^2=u\, , y^2=v$:
$$3(x^4+y^4)= 6x^2y^2\Leftrightarrow u^2+v^2=2uv \Leftrightarrow 0=u^2-2uv+v^2=(u-v)^2\Leftrightarrow u=v \Leftrightarrow x^2=y^2 \Leftrightarrow |x|=|y|$$
So $\det Df((x,y))=3(x^4+y^4)-6x^2y^2\neq 0$ and therefore invertable if $|x|\neq |y|$.
So for every $a\in U':=\{(x,y)\in\mathbb{R}^2|\, |x|\neq |y|\}$ there exists an open set $ U $ with $a\in U$ and an open set $V$ with $f(a)\in V$ such that $ f: U \rightarrow V $ is a local $C^1$-diffeomorphism $(*)$. So there exists a continuous differentiable inverse $ f^{-1}: V \to U $ .
Now I need to calculate $Df^{-1}(y)=[Df(a)]^{-1} $
How do i find the inverse of $Df((x,y))=\begin{pmatrix} 3x^2-y^2&-2xy\\2xy&x^2-3y^2 \end{pmatrix}$ with $|x|\neq |y|$? Gauss–Jordan elimination didn't work.