I am a total beginner in Riemannian geometry but I'm trying to teach myself the basics. So the following could contain many horrible mistakes.
Suppose I describe the x,y-plane by curvilinear coordinates $q_1$ and $q_2$:
$q_1: x^2-2-y = 0$
$q_2: -x^2+2-y =0$
Then the normalized tangent basis at each point along $q_1$ and $q_2$ is
$ \left\{ \frac{1}{1+4x^2} \begin{pmatrix} 1 \\ 2x \end{pmatrix}, \frac{1}{1+4x^2} \begin{pmatrix} 1 \\ -2x \end{pmatrix} \right\}$
No I simply use the classical dot product on the tangent basis for the metric tensor:
$g_{12} = \left< \frac{1}{1+4x^2} \begin{pmatrix} 1 \\ 2x \end{pmatrix},\frac{1}{1+4x^2} \begin{pmatrix} 1 \\ -2x \end{pmatrix} \right> = \frac{1-4x^2}{(1+4x^2)^2} = g_{21}$
And, equivalently $g_{11} = g_{22} = 1$.
Is that a correct definition of a Riemannian metric tensor?