Evaluate the surface integral $\iint_\Sigma (y^2-z^2)e^{yz}\,ds$ using Stokes' Theorem I found the following problem in a textbook:

Evaluate $$\iint_\Sigma (y^2-z^2)e^{yz}\,ds\,,$$ where $\;\Sigma\;$ is given by $x^2+y^2+z^2=1$, $\;z\geq0$, by evaluating the rotation of the field  $\;\mathbf{F}=(e^{yz}, 0, 0)$.

I evaluate the rotation to be $(0, ye^{yz}, -ze^{yz})$, and I also see that the integrand can be written $(0,y,z)\cdot \operatorname{rot}(\mathbf{F})$. From here I am not sure what to do next. Help appreciated!

$\vec F=(e^{yz}, 0, 0)$
So as you found, $\displaystyle \nabla \times \vec F = (0, ye^{yz}, -ze^{yz})$
Now the surface is $x^2+y^2+z^2 = 1, z \geq 0$. So the unit normal vector to the surface is $\hat n = (x, y, z)$.
So the surface integral is $\displaystyle \iint_\Sigma  (\nabla \times \vec F) \cdot \hat n \ dS = \iint_\Sigma  (y^2-z^2) e^{yz} \  dS$
which is what the question seeks us to evaluate.
The boundary of the surface is $x^2 + y^2 = 1, z = 0$. So using Stokes theorem, we can either convert it to another surface with the same boundary (read disk: $x^2+y^2 \leq 1, z = 0$) or we can convert it into line integral of $\vec F \ $ over boundary curve $x^2+y^2 = 1, z = 0$.
Choosing the latter, we parametrize the curve as
$r(t) =  (\cos t, \sin t, 0), 0 \leq t \leq 2\pi$
$r'(t) = (-\sin t, \cos t, 0)$
$\vec F = (e^{yz}, 0, 0) \implies \vec F (r(t)) = (1, 0, 0)$
So the line integral becomes,
$\displaystyle \int_0^{2\pi} (1, 0, 0) \cdot (- \sin t, \cos t, 0) \ dt = 0$
