Roughly, what we expect is that a single equation in three variables determines a surface in space; two equations determine a curve or curves (in the sense that the common solutions (x,y) of both equations form one or more curves); and three dermine a point or isolated points. Of course, exceptions abound: The solutions of the single equation x 2 + y 2 = 0 is the (one-dimensional) z-axis. And if we consider the infinitely many planes that all pass through the same line, then any two or more of their corresponding (linear) equations will still determine that common line. But "each new equation cuts down the dimension by one" is a handy rule of thumb.
Because I mentioned above that it is hard to see whether any points on the original surface x 2 z - xy 2 = 4 have the property that the normals there are parallel to the yz-plane, let us take one of the points on our curve and look at the gradient vector there, to see whether it has the desired properties.
One point on the surface is (-2,2,-1), and the gradient there is 8j + 4k, so it is clearly parallel to the yz-plane. The real question is the more basic one: Is it true that the gradient is always perpendicular to the contour at its base? Here is a piece of the original surface, with (-2,2,-1) at one corner and with the above gradient vector drawn in dark green.
The reader is invited to download the corresponding Winplot file and rotate it to see that gradient is indeed perpendicular to the surface.