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Jul 26, 2016 at 15:29 comment added Thomas Not at all : once you know a critical point $(x_0,y_0,z_0)$ you can think that the surface is given by $z(x,y)$. To compute second derivative $\partial ^2_x z (x_0,y_0,z_0)$, juts derive twice $f(x,y,z(x,y))$. Or derive once $f'_x(x,y,z(x,y))+f'_z(x,y,z) \partial _x z$, yiu get $f"_x(x_0,y_0,z_0)+f"_zz (x_0,y_0,z_0)\partial ^2_x z (x_0,y_0,z_0)=0$, as $ \partial _x z(x_0,y_0,z_0))=0$. What seems more tricky is to find easily the number of connected components
Jul 26, 2016 at 14:44 comment added Igor Rivin The index computation seems fairly horrible, a priori...
Jul 26, 2016 at 9:11 history answered Thomas CC BY-SA 3.0