Not as painful as it looks

Not nearly as painful as it looks, and a very clever question. Consider a closed surface formed by the given open surface, and the unit disk centered on the origin in the $xy$ plane. From the divergence theorem:

$\int \int_S \vec{F} \cdot \vec{n} \, dS = \int \int \int_V \nabla \cdot \vec{F} \, dV$

Since $F_x$ does not contain $x$ , $F_y$ does not contain $y$ , and $F_z$ does not contain $z$ , the divergence is zero. Therefore the net flux through the closed surface must be zero. A very useful consequence of this is that the fluxes through the unit disk and the defined surface must be equal and opposite. Consider a reduced $\vec{F}$ on the unit disk in the $xy$ plane, as well as the outward-facing normal vector (relative to the overall closed surface):

$\vec{F}_{reduced} = \Big( e^y, 0, \frac{3}{\pi} \Big ) \\ \vec{n} = (0,0,-1) \\ \int \int_{\text{unit disk}} \vec{F} \cdot \vec{n} \, dS = \int \int_{ud} -\frac{3}{\pi} dS = -\frac{3}{\pi} \int \int_{ud} dS = -\frac{3}{\pi} S = -\frac{3}{\pi} \, \pi = -3$

Since the flux associated with the upper surface cancels this, it must be $\boxed{3}$