### Legitimate coordinates?

Not good: $x,y$ and $-x,-y$have the same $x',y'$. Let us limit ourselves with $x>0$, $y>0$, then
$y=(x'y')^{1/2}, \quad x=(x'/y')^{1/2}\\ds^2=dx^2+dy^2= (\frac{\partial x}{\partial x'}dx' + \frac{\partial x}{\partial y'}dy')^2 + (\frac{\partial y}{\partial x'}dx'+ \frac{\partial y}{\partial y'}dy')^2\\=(\frac{1}{2\sqrt{x'y'}} dx'-\frac{\sqrt{x'}}{2\sqrt{{y'}^3}}dy')^2 + (\frac{\sqrt{y'}}{2\sqrt{x'}}dx' + \frac{\sqrt{x'}}{2\sqrt{y'}}dy')^2\\=\frac{1+{y'}^2}{4x'y'}{dx'}^2 + \frac{x'(1+y'^2)}{4{y'}^3}{dy'}^2 +\frac{1}{2}(\sqrt{x'y'}-\frac{\sqrt{x'}}{{y'}^2}) dx' dy'$