I struggle to come with an argument why any local metric on a Riemannian surface cannot be put into the form $$ g = \dfrac{d x^2}{F(x,y)} + F(x,y) d y^2. $$ so that the metric is diagonal, and the volume form is unit: $\omega = dx \wedge dy$.
I am aware of the following results (and their proofs), but they do not seem helpful in this case
- on a Riemannian surface one can always locally write metric as conformally flat (see isothermal coordinates on Wikipedia)
- on a Riemannian surface one can always locally diagonalize the metric with the given function being one of the coordinates (see, e.g. Appendix C in arxiv.org/abs/hep-th/0408141)
As @MikhailKatz pointed in the comments, the only local obstruction to introduction of a metric on a Riemannian surface is a Gaussian (or scalar) curvature. The latter is given by $$ R = -F_{xx} + F^{-3} \left( F F_{yy} - 2 F^2_{y}\right) $$ where lower indices stand for respective derivatives, for example, $F_{xx} = \partial_x^2 F(x,y)$.
If one treats the equation as an inhomogeneous PDE for $F$, then by Cauchy-Kowalewskaya theorem, for any analytic $R$ one can always supply boundary conditions such that the solution to the boundary problem exists.
P.S. This is a very sketchy proof I made to convince myself. Please point out if my application of C-K theorem was incomplete, or the result can be strengthened to smooth functions too.