Triangulation of vertices of an ellipsoid
If you sample the two parameters $\eta$ and $\omega$ with steps $d\eta$ and $d\omega$, then you'll get a grid of points $v_{ij} = f(i\;d\eta,j\;d\omega)$.  Any four adjacent points will define a quadrilateral.  To get triangles, you just have to split each quad in two by a diagonal.



So in the example, you'd split the quadrilateral $\{v_{00},v_{01},v_{11},v_{10}\}$ into two triangles $\{v_{00},v_{01},v_{10}\}$ and $\{v_{01},v_{11},v_{10}\}$.  You'll want to make sure the points come in the right order so the triangle orientations match (or you'll end up backface-culling half of your triangles.)

Your second equation, for $n_{xyz}$, is finding the normal of each vertex, which you can use for shading.