Answer
The curve belongs to the cone $x^{2}+y^{2}=z^{2}$.
Work Step by Step
Squaring all three parametric equatins, we find
$x^{2}=(1+\cos 16t)^{2}\cos^{2}t,$
$y^{2}=(1+\cos 16t)^{2}\sin^{2}t,$
$z^{2}=(1+\cos 16t)^{2}.$
So, for any point (x,y,z) on the curve, it follows that
$x^{2}+y^{2}=(1+\cos 16t)^{2}(\cos^{2}t+\sin^{2}t)=(1+\cos 16t)^{2}=z^{2}.$
The point also belongs to the cone defined with
$x^{2}+y^{2}=z^{2}$.
(Graphed with Geogebra)
