Of course, you will have to do upper and lower hemispheres separately. One way to get the projection into the xy-plane is to find the gradient of x2+ y2+z2, 2xi+ 2yj+ 2zk, and "normalize by dividing by 2z: (x/z)i+ (y/z)j+ k. Then n dS is (x/z)i+ (y/z)j+ k dxdy.
f.n dS is ((x2/z)+ (y2/z)- 2z) dxdy. I think I would rewrite that as ((x2/z)+ (y2/z)+ z- 3z) dxdy= ((x2+ y2+ z2)/z- 3z) dxdy= (a^2/z- 3z)dxdy. Now, for the upper hemisphere, [itex]z= \sqrt{a^2- x^2- y^2}[/itex] while for the negative hemisphere it is the negative of that. Because your integrand is an odd function of z, I think the symmetry of the sphere makes this obvious.
Finally, do you know the divergence theorem?
[tex]\int\int_T\int (\nabla \cdot \vec{v}) dV= \int\int_S (\vec{v} \cdot \vec{n}) dS[/tex]
where S is the surface of the three dimensional region T. Here [itex]\nabla\cdot f[/itex] is very simple and, in fact, you don't have to do an integral at all! I wouldn't be surprized to see this as an exercise in a section on the divergence theorem.
)