Answer
The flux is
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = - 4$
Work Step by Step
The ellipsoid ${\left( {\frac{x}{4}} \right)^2} + {\left( {\frac{y}{3}} \right)^2} + {\left( {\frac{z}{2}} \right)^2} = 1$ can be parametrized by
$G\left( {\theta ,\phi } \right) = \left( {4\cos \theta \sin \phi ,3\sin \theta \sin \phi ,2\cos \phi } \right)$
So,
${{\bf{T}}_\theta } = \left( { - 4\sin \theta \sin \phi ,3\cos \theta \sin \phi ,0} \right)$
${{\bf{T}}_\phi } = \left( {4\cos \theta \cos \phi ,3\sin \theta \cos \phi , - 2\sin \phi } \right)$
${\bf{N}}\left( {\theta ,\phi } \right) = \left| {\begin{array}{*{20}{c}}
{\bf{i}}&{\bf{j}}&{\bf{k}}\\
{ - 4\sin \theta \sin \phi }&{3\cos \theta \sin \phi }&0\\
{4\cos \theta \cos \phi }&{3\sin \theta \cos \phi }&{ - 2\sin \phi }
\end{array}} \right|$
$ = \left( { - 6\cos \theta {{\sin }^2}\phi } \right){\bf{i}} - \left( {8\sin \theta {{\sin }^2}\phi } \right){\bf{j}} + \left( { - 12\cos \phi \sin \phi } \right){\bf{k}}$
Since $x,y,z \le 0$, the portion of $S$ is in the range: $\pi \le \theta \le \frac{{3\pi }}{2}$ and $\frac{\pi }{2} \le \phi \le \pi $. In this range the $z$-component of ${\bf{N}}\left( {\theta ,\phi } \right)$ is always positive, thus it is upward-pointing normal.
So,
${\bf{F}}\left( {G\left( {\theta ,\phi } \right)} \right) = 2\cos \phi {\bf{i}}$
${\bf{F}}\left( {G\left( {\theta ,\phi } \right)} \right)\cdot{\bf{N}}\left( {\theta ,\phi } \right) = - 12\cos \theta \cos \phi {\sin ^2}\phi $
We evaluate the flux of ${\bf{F}} = z{\bf{i}}$ over the portion of $S$ using Eq. (3):
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = \mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\left( {G\left( {\theta ,\phi } \right)} \right)\cdot{\bf{N}}\left( {\theta ,\phi } \right){\rm{d}}\theta {\rm{d}}\phi $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = - 12\mathop \smallint \limits_{\theta = \pi }^{3\pi /2} \mathop \smallint \limits_{\phi = \pi /2}^\pi \cos \theta \cos \phi {\sin ^2}\phi {\rm{d}}\theta {\rm{d}}\phi $
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = - 12\mathop \smallint \limits_{\theta = \pi }^{3\pi /2} \cos \theta {\rm{d}}\theta \mathop \smallint \limits_{\phi = \pi /2}^\pi \cos \phi {\sin ^2}\phi {\rm{d}}\phi $
Write $t = \sin \phi $. So, $dt = \cos \phi d\phi $. The integral becomes
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = - 12\left( {\sin \theta |_\pi ^{3\pi /2}} \right)\mathop \smallint \limits_{t = 1}^0 {t^2}{\rm{d}}t$
$\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = - 12\left( { - 1} \right)\left( {\frac{1}{3}{t^3}} \right)|_1^0 = - 4$
So, the flux is $\mathop \smallint \limits_{}^{} \mathop \smallint \limits_S^{} {\bf{F}}\cdot{\rm{d}}{\bf{S}} = - 4$.
