I would like to know the reasons why the existance of Chebyshev net in 3D-case is problematic. This question boils down to the PDE described below. (I do not know much about PDEs, so feel free to say something trivial.)
Set $$\mathbb W^2 =\{\\,(x,y,z)\in \mathbb R^3\mid x+y+z=0\\,\}.$$$$\mathbb W^2 =\{\,(x,y,z)\in \mathbb R^3\mid x+y+z=0\,\}.$$ Given a smooth map $f: \mathbb W^2\to \mathbb R^3$ which is $C^\infty$-close to the identity map, I need to extend $f$ to a neighborhood of $\mathbb W^2$ so that it satisfies the following three equations: $$\left|\frac{\partial f}{\partial x}\right|^2=\left|\frac{\partial f}{\partial y}\right|^2=\left|\frac{\partial f}{\partial z}\right|^2=1.$$
Comments.
The 2D-case has a solution, this is due to Chebyshev and it is more than 100 years old.
In general, 3D Chebyshev net is a solution of 3 equasions as above where $f$ maps a domain in $\mathbb R^3$ in a 3-dimensional Riemannian manifold and $|{*}|$ is defined by its metric. I guess this case is just as hard as the special case described above.
Assuming that $x+y+z>0$ and $f(x,y,z)$ is defined, it is easy to see that $f(x,y,z)$ depends only on the values $f(x',y',z')$ for $(x',y',z')$ in the triangle $\Delta\subset\mathbb W^2$ defined by the inequalities $$x'\le x,\ y'\le y,\ z'\le z.$$