The square (call it [tex]S[/tex]) has one vertex at the origin (0, 0, 0) and one edge on the y-axis, which tells us another vertex is (0, 3, 0). The normal vector to the plane is [tex]\vec n=\vec\imath-\vec k[/tex], which is enough information to figure out the equation of the plane containing [tex]S[/tex]:
[tex](x\,\vec\imath+y\,\vec\jmath+z\,\vec k)\cdot(\vec\imath-\vec k)=0\implies x-z=0\implies z=x[/tex]
We can parameterize this surface by
[tex]\vec s(x,y)=x\,\vec\imath+y\,\vec\jmath+x\,\vec k[/tex]
for [tex]0\le x\le\frac3{\sqrt2}[/tex] and [tex]0\le y\le3[/tex]. Then the flux of [tex]\vec F[/tex], assumed to be
[tex]\vec F(x,y,z)=(e^{xy}+9z+4)\,\vec\imath+(e^{xy}+4z+9)\,\vec\jmath+(9ze^{xy})\,\vec k[/tex],
is
[tex]\displaystyle\iint_S\vec F(x,y,z)\cdot\mathrm d\vec S=\iint_S\vec F(\vec s(x,y))\cdot\vec n\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle\int_0^3\int_0^{3/\sqrt2}\left((4+e^{xy}+9x)\,\vec\imath+(9+e^{xy}+4x)\,\vec\jmath+(e^{xy}+9x)\,\vec k\right)\cdot(\vec\imath-\vec k)\,\mathrm dx\,\mathrm dy[/tex]
[tex]=\displaystyle\int_0^3\int_0^{3/\sqrt2}4\,\mathrm dx\,\mathrm dy=\boxed{18\sqrt2}[/tex]
