I assume [tex]C[/tex] has counterclockwise orientation when viewed from above.
By Stokes' theorem,
[tex]\displaystyle\int_C\vec F\cdot\mathrm d\vec r=\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S[/tex]
so we first compute the curl:
[tex]\vec F(x,y,z)=xy\,\vec\imath+yz\,\vec\jmath+xz\,\vec k[/tex]
[tex]\implies\nabla\times\vec F(x,y,z)=-y\,\vec\imath-z\,\vec\jmath-x\,\vec k[/tex]
Then parameterize [tex]S[/tex] by
[tex]\vec r(u,v)=\cos u\sin v\,\vec\imath+\sin u\sin v\,\vec\jmath+\cos^2v\,\vec k[/tex]
where the [tex]z[/tex]-component is obtained from
[tex]1-(\cos u\sin v)^2-(\sin u\sin v)^2=1-\sin^2v=\cos^2v[/tex]
with [tex]0\le u\le\dfrac\pi2[/tex] and [tex]0\le v\le\dfrac\pi2[/tex].
Take the normal vector to [tex]S[/tex] to be
[tex]\vec r_v\times\vec r_u=2\cos u\cos v\sin^2v\,\vec\imath+\sin u\sin v\sin(2v)\,\vec\jmath+\cos v\sin v\,\vec k[/tex]
Then the line integral is equal in value to the surface integral,
[tex]\displaystyle\iint_S(\nabla\times\vec F)\cdot\mathrm d\vec S[/tex]
[tex]=\displaystyle\int_0^{\pi/2}\int_0^{\pi/2}(-\sin u\sin v\,\vec\imath-\cos^2v\,\vec\jmath-\cos u\sin v\,\vec k)\cdot(\vec r_v\times\vec r_u)\,\mathrm du\,\mathrm dv[/tex]
[tex]=\displaystyle-\int_0^{\pi/2}\int_0^{\pi/2}\cos v\sin^2v(\cos u+2\cos^2v\sin u+\sin(2u)\sin v)\,\mathrm du\,\mathrm dv=\boxed{-\frac{17}{20}}[/tex]
