Got a reply from the OP:
Actually I have devoted hours and hours to this problem. It is in Chapter 9 of the old Schaum's book. I have tried to put this in a form u*e1 + v*e2 + f(u,v)*e3 with little success. I have used other hints from this chapter such as determinant [dx, fx, Dfx] = 0 and I can get some simplifications but then I get lost in overly extensive equations. I have tried graphing this function as z = arctan(y/x) but without Mathematica or Maple the graph gets hazy in my mind. I know it would have several sheets but try to stick with the "main sheet". So I feel I am missing something crucial, but at the same time feel it should have been easy and that it is a really cool problem.
Decided to post that here along with my help for future reference, in case other people google his question and arrive at this page.
We can rewrite (as noted by the OP) the equation to x tan(z) = y.
Using this, we can parametrize the surface as follows:
x(u,v) = (u, u \tan v, v)
Note: x is now not the first coordinate, but rather the position vector
The only economical way I know of calculating the principal values, is using the formulae k_{1,2} = H \pm \sqrt{H^2 - K}.
So we need to calculate H and K. This can be easily done using the parametrization of the surface and the following:
H \propto En - 2Fm + Gl \qquad \qquad K = \frac{ln-m^2}{EG-F^2}
where E:= x_u \cdot x_u \qquad F:= x_u \cdot x_v \qquad G = x_v \cdot x_v
and l := x_{uu} \cdot \xi \qquad m := x_{uv} \cdot \xi \qquad n := x_{vv} \cdot \xi
where xi is the normal \xi = x_u \times x_v.
I've only given H up to something proportional to it, because you'll only have to prove that H = 0, which you can expect by comparing the formulae for k_1 and k_2 given above with what you know will be the results.