If $f:\mathbb{R}^2\to\mathbb{R}^2, f(x,y)=(x\cos (y), \sin (x-y))$. Then $\exists X,Y\subset\mathbb{R}^2$ both open such that $(\pi/2,\pi/2)\in X$ and $f:X\to Y$ is a diffeomorphism.
I'm not sure how can a diffeomorphism be defined with the condition $(\pi/2,\pi/2)\in X$. Because if $f$ diffemoprhism can be defined then $Jf(a)\neq0 \;\forall a\in X$ but
$\det Jf(a)=\det\begin{pmatrix} \cos y & -x\sin y\\ \cos(x-y) & -cos(x-y)\end{pmatrix} = \cos(x-y)(x\sin y - \cos y)$
Then must be:
(1) $\cos(x-y)\neq 0 \implies x-y\neq \frac{1}{2}\pi+k\pi$. (2) $x\sin y - \cos y \neq 0 \implies x\neq\cot y$ if $y\neq k\pi$ (and this inequality must hold for the equation because if $y=k\pi$ then $x\sin y-\cos y \neq 0)$.
With (1) and (2) I have that I need $(x,y)$ such that $y+ \pi(k+\frac{1}{2})\neq\cot y$ but I have no clue what could I do from here, this is, how to find solutions to the last trig. equation.
Another idea was to get rid of what I did before and just set $(\pi/2,\pi/2)$ as the center of a neighborhood but to set a useful radius I still would need which points make $\det Jf(a)=0$ in order to avoid them, and this would lead me again to the mess I wrote before -(1) and (2)-.
