Consider the transformation $$x=u^2_1-u_2^2$$ $$y=2u_1u_2$$ $$z=u_3$$
Show that $(u_1,u_2,u_3)$ form right-handed orthogonal curvilinear coordinates.
My issue is expressing x,y in terms of $u_1,u_2$. I have tried many different combinations to try to get an explicit answer, like $x^2+y^2$ and $x^2-y^2$, and doing different substitutions, but I my answer is always messy and implicit.
Might be helpful to use the definitions
$$\mathbf e_{u_i} = \frac{1}{\alpha_i} \bigg(\frac{\partial x}{\partial u_i} \mathbf e_x + \frac{\partial y}{\partial u_i} \mathbf e_y + \frac{\partial x}{\partial u_i} \mathbf e_z\bigg) \qquad i=1,2,3$$
where $\alpha_i$ are the normalising constants.
Computing the partial derivatives, we find that
\begin{align} \mathbf e_{u_1} & \propto 2u_1\mathbf e_x + 2u_2\mathbf e_y \\ \mathbf e_{u_2} & \propto -2u_2\mathbf e_x + 2u_1\mathbf e_y \\ \mathbf e_{u_3} & \propto \mathbf e_z \end{align}
Normalising, we obtain
\begin{align} \mathbf e_{u_1} & = \frac{u_1\mathbf e_x + u_2\mathbf e_y}{\sqrt{u_1^2+u_2^2}} \\ \mathbf e_{u_2} & = \frac{-u_2\mathbf e_x + u_1\mathbf e_y}{\sqrt{u_1^2+u_2^2}} \\ \mathbf e_{u_3} & = \mathbf e_z \end{align}
It is then easy to check that
$$\mathbf e_{u_1} \times \mathbf e_{u_2} = \mathbf e_{u_3} \qquad \mathbf e_{u_2} \times \mathbf e_{u_3} = \mathbf e_{u_1} \qquad \mathbf e_{u_3} \times \mathbf e_{u_1} = \mathbf e_{u_2}$$