What Stokes' Theorem tells you is the relation between the line integral of the vector field over its boundary $\partial S$ to the surface integral of the curl of a vector field over a smooth oriented surface $S$:
$$\oint\limits_{\partial S} \mathbf{F} \cdot d\mathbf{r} = \iint\limits_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \tag{1}\label{1}$$
Since the prompt asks how to calculate the integral using Stokes' Theorem, you can find a good parametrization of the boundary $\partial S$ and calculate the "easier" integral of the LHS of (1).
Note that the boundary of $S$ is given by:
$$\partial S=\{(x,y,z)\in \mathbb R^3 : x^2 +y^2 =25, z=0\},$$
so a good parametrization to use for $\partial S$ could be:
$$\sigma :[0,2\pi] \subseteq \mathbb{R} \rightarrow \mathbb R^3$$ $$\sigma(\theta)=(5\cos(\theta),5\sin(\theta),0)$$
and finally the integral to calculate ends up being:
$$\begin{align}\oint\limits_{\partial S} \mathbf{F} \cdot d\mathbf{r}&= \iint\limits_S (\nabla \times \mathbf{F}) \cdot d\mathbf{S} \\ &=\int_{0}^{2\pi}(0,5\cos(\theta),e^{25\cos(\theta)\sin(\theta)})\cdot(-5\sin(\theta),5\cos(\theta),0)\, d\theta,\\&=\int_{0}^{2\pi}25\cos^2(\theta)\,d\theta\end{align}$$
and from here you can use trigonometric identities to calculate the last integral.
I hope that helps!
In a sense, Stokes', Green's, and Divergence theorems are all special cases of the generalized Stokes theorem for differential forms $$\int_{\partial \Omega} \omega = \int_\Omega d\omega$$
but I don't think that's what you're asking about.
The usual (3-dimensional) Stokes' and Divergence theorems both involve a surface integral, but they are in rather different circumstances.
In the Divergence theorem, the surface $S$ is the boundary of a bounded region $R$ of space, and you're taking the flux through this surface of a vector field $\bf F$ defined in $R$ and on its boundary:
$$ \iint_S {\bf F} \cdot d{\bf S} = \iiint_R \text{div}\;{\bf F}\; dV $$
In Stokes' theorem, the surface is generally not the boundary of a region: instead it has a boundary which is a curve $C$; you're taking the flux, not of an arbitrary vector field, but of the curl of some other field:
$$ \iint_S \text{curl}\; {\bf G} \cdot d{\bf S} = \oint_C {\bf G} \cdot d{\bf r} $$
There is one situation where both apply: suppose your surface $S$ is the boundary of a bounded region $R$, and your vector field $\bf F$
happens to be the curl of some other field $\bf G$. Since the divergence of the curl is $0$, the Divergence theorem says the result is $0$. On the other hand, for Stokes the surface has no boundary (it's a closed surface), so Stokes integrates $\bf G$ around an empty curve and gets $0$ as well.
Best Answer
The result of the line integral is correct, so the mistake must be with your calculation of the surface integral. You have $\nabla \times \vec F = \left( 0,1-2xz,y^2 \right)$ and with a surface of the form $z=g(x,y)$, the normal $\vec n$ is given by $\left(-g_x,-g_y,1\right)$. The projection of the surface onto the $xy$-plane is a disc centered in the origin and with radius $\sqrt{6}$, so you have to integrate over this disc $D$: $$\iint_D \left( 0,1-2xz,y^2 \right) \cdot \left(-g_x,-g_y,1\right) \,\mbox{d}A$$ where $z=g(x,y)$ with $g(x,y)=x^2+y^2$, so: $$\iint_D \left( 0,1-2x\left(x^2+y^2\right),y^2 \right) \cdot \left(-2x,-2y,1\right) \,\mbox{d}A = \iint_D \left( 4 x^3 y + 4 x y^3 + y^2 - 2 y \right) \,\mbox{d}x\,\mbox{d}y$$ You can choose to switch to polar coordinates or not, but you should find $9\pi$ either way.
Maybe this is sufficient to find your mistake? If not, perhaps you can show us your calculations.