No. f(x,y,z) is a more general case of z=f(x,y).
Consider 2 dimensions first. y=f(x) gives a curve with a few interesting properties, one of which is called one-to-one correspondence. That means that you can draw a vertical line anywhere on your graph, and it will cross the curve exactly once. (The same need not be true for a horizontal line.)
If I relax that restriction and re-arrange the equation, I can write f(x,y)=const (still in 2D here). This still gives me a curve, but I don't have the one-to-one correspondence anymore.
For example, x^2 + y^2 = 1 is the equation for a unit circle. That is, all the (x,y) pairs which satisfy that equation lie on a circle centered at origin with unity radius.
Now extend the thinking to 3D. Unsurprisingly, x^2 + y^2 + z^2 = 1 is the equation for a sphere centered at origin and with unity radius. This is an equation in 3 variables of the form f(x, y, z) = const. If we were to rewrite that equation in the form z = f(x, y), we would see that f(x, y) = sqrt(x^2 + y^2). This equation has two solutions; one for the "top half" of the sphere, and one for the "bottom half". That is because the square root function is ambiguous: say y=sqrt(x), then x^2=y, but also (-x)^2=y.
So it is the same difference in 2D and 3D. You can draw a line anywhere, so long as it's parallel to the z-axis and it will intersect the surface exactly once in the case of z=f(x,y). No so if f(x,y,z)=const. We relax the correspondence rule and can get all sorts of wild shapes.