-x^2 - y^2 - 16z^2= 0
how do i go about graphing this? all i know is it has to do with looking at "sections"
The general equation for an ellipsoid is $\displaystyle \frac{(x-h)^2}{a^2}+\frac{(y-k)^2}{b^2}+\frac{(z-l)^2}{c^2}=1$, where $\displaystyle (h,k,l)$ is the center of the ellipsoid, and $\displaystyle a>0,\,b>0,\,c>0$.
Unfortunately, the equation $\displaystyle -x^2-y^2-16z^6=0$ isn't an ellipsoid. I'm not even sure what to call it since the degree of z is 6... Are you sure you copied it down right?
Ok, that's a little better. Are you sure its not $\displaystyle -x^2-y^2{\color{red}+}16z^2=0$? Because that would give us an elliptic [circular] cone!
Note that $\displaystyle -x^2-y^2+16z^2=0\implies z^2=\frac{x^2}{16}+\frac{y^2}{16}$, which is an elliptic cone centered at (0,0,0).
The graph would be as follows: