Show that the set:
$ S=(\{ (x,y,z) \in \mathbb{R}^{3}; (x^2+y^2+z^2)^2 = a^2x^2+b^2y^2+c^2z^2 \} - \{(0,0,0) \} )$
is a regular surface.
I've tried use the regular value theorem with $f(x,y,z)=(x^2+y^2+z^2)^2-a^2x^2-b^2y^2-c^2z^2$
Finding the partial derivatives:
$f_x=4x(x^2+y^2+z^2)-2a^2x, f_y=4y(x^2+y^2+z^2)-2b^2y,f_z=4z(x^2+y^2+z^2)-2c^2z$
I need to get to that $0$ is a regular value, but i don't know how to show that because $f_x(0)=f_y(0)=f_z(0)=0$.
I appreciate your collaboration.