f(x,y)=9-x2-9y2 is based on the ellipse, that is, when f(x,y)=0, it's the base resting on the x-y plane.
x2+9y2=9 which is x2/9+y2=1, the semi-major axis has length 3, and the semi-minor axis length 1.
When f(x,y)=5, x2+9y2=4, that is, x2/4+9y2/4=1, these semi-axes have length 2 and ⅔ respectively.
When f(x,y)=9, the ellipse collapses to a point (0,0,9), which is the highest point of the dome.
(This can be viewed in 3D using filtered glasses. Note the differences in scale on the axes: y is graduated in 0.5 units, 0.5, 1.5, 2, ... while x is graduated in single units, 1, 2, 3, ... and z is graduated in double units, 2, 4, 6, ... so as to show the dome clearly.)
If we write the equation as x2+9y2=9-f(x,y), x=√(9-f(x,y)-9y2), so the domain is constrained to:
9-f(x,y)-9y2≥0, that is, when f(x,y)=9, x=y=0. The domain is all points on the surface of the elliptical dome, which continues indefinitely below the x-y plane.
More specifically, since f(x,y) = 9-x2-9y2, where the RHS is the equation of an ellipse, then the domain of f must be wherever 9-x2-9y2 can be calculated. Let z=f(x,y).
When x=0, z=f(0,y)=9-9y2, which is a parabola in the z-y plane (inverted U shape). Similarly when y=0, z=f(x,0)=9-x2, which is a wider parabola in the z-x plane. This can be seen from the picture. There is no restriction on either x or y although z is limited. The surface of the elliptical dome is unbounded, hence the domain is unbounded.