I will show my question with the help of an exercise I stumbled upon recently because I think it will illustrate my problem more clearly. I'm simplifying the exercise by just showing the steps and not the actual problem because it is just for illustration purposes.
We start in the cartesian coordinate system, then there is a coordinate change.
$x\rightarrow v, x =e^v -1 \Rightarrow dx = e^vdv, dx^2 = e^{2v}dv^2 $
$y \rightarrow w, y = w \Rightarrow dy = dw, dy^2= dw^2$
For a point P(4,3) in (x,y):
$v = ln(x+1) = ln(5) = 1.6 $ and $ w = y = 3$
$P(4,3) \rightarrow P'(1.6,3)$
The distance to the point P now has to be calculated.
In the (x,y) coordinates:
$ds^2 = dx^2 + dy^2 = 4^2 + 3^2 = 16 + 9 = 25$
In the (v,w) coordinates:
$ds^2 = e^{2v}dv^2 + dw^2 = e^{2*1.6}*1.6^2 + 3^2 = 71.8$
Isn't $ds^2$ supposed to be invariant following $ds^2 = g_{\mu\nu}dx^{\mu}dx^{\nu}$ ? The metric tensor has changed because of the coordinate change, and therefore the value of $ds^2$ has changed. Is that right ? If so, then in what way is $ds^2$ invariant ?