= Solution
For $h=x^2/a^2+y^2/b^2+z^2-1$, the <gradient> $\nabla h=2(x/a^2,y/b^2,z)$ is a <normal vector> to the <ellipsoid>. The outward <unit normal vector> is $(x/a^2,y/b^2,z)/\sqrt{x^2/a^4+y^2/b^4+z^2}$. The conditions $a,b>\sqrt2$ ensure that $z=\sqrt{1-x^2/a^2-y^2/b^2}>0$ everywhere on the rectangular projection.
The <dot product> $F\cdot\nabla h=2xy(1/b^2-1/a^2)$ vanishes pointwise when $a=b$. In that case $F$ has no normal component, so \b[the <surface flux> is zero]. For unequal axes, the given <vector surface element> gives
$$
\Phi=\left(\frac1{b^2}-\frac1{a^2}\right)\int_0^1\int_0^1\frac{xy}{\sqrt{1-x^2/a^2-y^2/b^2}}\,dy\,dx.
$$
Put $u=x^2/a^2$, $v=y^2/b^2$. Then $xy\,dxdy=a^2b^2\,dudv/4$, and
$$
\Phi=\frac{a^2-b^2}{4}\int_0^{1/a^2}\int_0^{1/b^2}(1-u-v)^{-1/2}\,dv\,du.
$$
Two elementary <integrations> now give
$$
\boxed{\Phi=\frac{a^2-b^2}{3}\left[1-\left(1-\frac1{a^2}\right)^{3/2}-\left(1-\frac1{b^2}\right)^{3/2}+\left(1-\frac1{a^2}-\frac1{b^2}\right)^{3/2}\right].}
$$
This is the <rotational flux through an ellipsoidal graph patch> with unit vertical semiaxis and unit projection lengths. It also tends continuously to zero when $a=b$, agreeing with the pointwise argument.
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