If I have a smooth surface $M$ (2D embedded in 3D), under what conditions I can assure that there exists a finite collection of charts $\{U\_i, \phi\_i \}\_{i}$, with $\phi\_i : U\_i \to M$, such that its directional derivatives have always the same magnitud and that they are orthogonal set of coordinates?, i.e.


$$
|\partial\_x \phi\_i| = |\partial\_y \phi\_i| \\
\partial\_x \phi\_i \cdot \partial\_y \phi\_i = 0
$$


Trivially flat surfaces have this property, and I suspect that is the general case, but I'm not really sure.



Requirement of parametrization of surfaces
You can always do this, but it's not as simple as using some kind of ODE (such a flow of vector fields) to construct such charts.


First, assume that your surface is connected and simply-connected. Then it's either compact, in which case it's a $2$-sphere, or it's diffeomorphic to $\mathbb{R}^2$. In the compact, case, the uniformization theorem implies that it's conformally equivalent to the standard $2$-sphere, i.e., there is a smooth diffeomorphism $f:S^2\to M$ such that the $f$-pullback of the induced metric on $M$ is a multiple of the standard metric on $S^2$. Now composing $f$ with the inverses of the usual stereographic projections (which are conformal) of the sphere to the plane from the north and south poles gives you two coordinate patches on $M$ that are conformal. In the non-compact case, the uniformization theorem says that there is a single smooth diffeomorphism $\phi:U\to M$ that is conformal, i.e., the $\phi$-pullback of the induced metric on $M$ is a multiple of the standard metric on $U\subset\mathbb{R}^2$. Here, by the Riemann mapping theorem, we can either take $U$ to be $\mathbb{R}^2$ or to be the unit disk.


To cover the general case, you only need to know that $M$ has a finite open cover by simply-connected open sets. This is a standard fact about 2-dimensional surfaces.


