What? HOW do you get that?
First, z= 3x2+ 2y2+ 1 involves three variables, x, y, and z, and so is a three-dimensional graph. It is, simply, a "paraboloid". It may be that your "two parabolas" are when you take x= 0 and then y= 0. If so, that's not a bad way to start. Imagine, with x= 0, that you are drawing the parabola in the yz-plane. Then with y= 0, you are drawing the parabola in the xz-plane. If you draw your x-z axes on the paper and imagine the y-axis coming out of the paper, then your two parabolas are at right angle to each other. Imagine the full graph rotating around the z-axis to meet those two parabolas.
As for getting a line for [itex]x^2+ y^2\le 1[/itex], I can't make heads of tails out of that! you should know that the graph of [itex]x^2+ y^2= 1[/itex] is a circle with center at (0,0) and radius 1. The graph of [itex]x^2+ y^2\le 1[/itex] is all points on or inside that circle- a "disk".