Answer
Hyperboloid of two sheets; the trace is an ellipse in the $ xy $-plane.
Work Step by Step
The equation
$$
4x^2+ \left(\frac{y}{3}\right)^{2} -2z^2=-1
$$
can rewritten as follows
$$
\left(\frac{x}{1/2}\right)^{2}+ \left(\frac{y}{3}\right)^{2} =\left(\frac{z}{1/\sqrt 2}\right)^{2}-1
$$
which is a hyperboloid of two sheets.
(See equations on page 691.)
To find the trace with the plane $ z=1$, we have
$$
\left(\frac{x}{1/2}\right)^{2}+ \left(\frac{y}{3}\right)^{2} =1
$$
which is an ellipse in the $ xy $-plane.
