I was told that the line element in ellipsoidal coordinates is: $$ds^2=a^2(u^2+v^2)\bigg(\frac{du^2}{u^2+1}+\frac{dv^2}{1-v^2}\bigg)+a^2(u^2+1)(1-v^2)d\phi^2$$
I've been trying to obtain this using $ds^2=x^2+y^2+z^2$, and the coordinate transformation from cartesian to ellipsoidal given by:
$$\begin{align} x^2 &= (a^2+\xi)(a^2+\eta)(a^2+\zeta)/(b^2-a^2)(c^2-a^2)\\ y^2 &= (b^2+\xi)(b^2+\eta)(b^2+\zeta)/(a^2-b^2)(c^2-b^2)\\ z^2 &= (c^2+\xi)(c^2+\eta)(c^2+\zeta)/(a^2-c^2)(b^2-c^2). \end{align}$$
The problem is that I don't recover the same line element and I can't find any source of how to calculate this $ds^2$. I think maybe they just used a simplified version of these coordinates or maybe they named the coordinates wrong. Anyone can identify such line element? or there is any subtlety for calculating it?
You can just find the metric tensor $g_{\mu\nu}$. Let $\xi_1,\xi_2,\xi_3$ be the coordinates to which you are trying to transform, and let $x_1,x_2,x_3=x,y,z$. Then, the entries of the metric tensor are given by
\begin{align} g_{\mu\nu}=\sum_\alpha\frac{\partial\xi_\alpha}{\partial x_\mu}\frac{\partial\xi_\alpha}{\partial x_\nu}. \end{align}
You should get a $3\times 3$ matrix. Then, the line element in ellipsoidal coordinates is given by
\begin{align} ds^2=\sum_\mu\sum_\nu g_{\mu\nu}dx_{\mu} dx_{\nu}. \end{align}