Question: I need convert this spherical expression, to a rectangular form (specific surface). $$\rho^2\cos(2\phi)-1=0$$
Thanks for a while.

Answer: 
If $\phi$ is the polar and $\theta$ the azimuthal coordinate, using double angle trigonometric identity your surface is described by $$\rho^2\cos(2\phi)=\rho^2\cos^2(\phi)-\rho^2\sin^2(\phi)=1,$$i.e. $$\rho^2\cos^2(\phi)-\rho^2\sin^2(\phi)(\sin^2\theta+\cos^2\theta)=1$$
which transforming to Cartesian coordinates yields the hyperboloid
$$z^2-(y^2+x^2)=1$$.
$(x^2+y^2)\cos(2\arctan(y/x))-1 = (x^2+y^2).\frac{x^2-y^2}{x^2+y^2} -1 = x^2-y^2-1$