When you are calculating a flux integral over a surface ##S##$$
\iint_S \vec F\cdot d\vec S=\iint_S\vec F \cdot \hat n dS$$you parameterize the surface in convenient variables ##u## and ##v## (which may be just ##x## and ##y##)$$
\vec R = \vec R(u,v)$$The normal to the surface is ##\pm \vec R_u\times \vec R_v##, with the sign chosen to agree with the orientation. Then to get a unit normal you have$$
\hat n =\pm \frac{\vec R_u\times \vec R_v}{|\vec R_u\times\vec R_v|}$$The scalar ##dS\ ##is given by$$
dS=|\vec R_u\times \vec R_v|dudv$$When you put these in the formula for the flux integral you get$$
\iint_S \vec F\cdot d\vec S=\iint_S\vec F \cdot \hat n dS=
\pm\iint_{(u,v)}\vec F\cdot \frac{\vec R_u\times \vec R_v}{|\vec R_u\times\vec R_v|}
|\vec R_u\times \vec R_v|dudv
=\pm \iint_{(u,v)}\vec F\cdot \vec R_u\times \vec R_vdudv$$where the limits are for the region in the ##u-v## plane. Note that you never have to calculate ##|\vec R_u\times\vec R_v|##. In your problem what he means by parameterizing the surface is writing it as ##\vec R(x,y) = \langle x,y,9-x^2-y^2\rangle##. If you calculate ##\vec R_x\times \vec R_y## you will get ##\langle 2x,2y,1\rangle##, which happens to have ##z## positive so agrees with your orientation, and agrees with your gradient. You have calculated ##\nabla \times \vec F = \langle -4,6,-3\rangle##, which is the ##\vec F## to put in the formula above:$$
\iint_S \langle -4,6,-3\rangle\cdot d\vec S = +\iint_{(x,y)}\langle -4,6,-3\rangle\cdot
\langle 2x,2y,1\rangle dydx=\iint_{(x,y)}-8x+12y-3\ dydx$$where the ##x-y## limits are over the region in the ##xy\ ##plane. That is where ##z=0## and gives the region enclosed by ##x^2+y^2=9##. At this point, you might want to work the integral in polar coordinates.
Then, of course, you still have to work the other side of the Stokes Theorem equation and see you get the same answer.