Prove $ \int_{C}fdr=\int_{S}dS\times\nabla f$ Prove $\displaystyle \int_{C}fdr=\int_{S}dS\times\nabla f$.
where $C=\partial S$ and the usual relationship between orientations hold.
Apply Stokes's theorem to $F=af$ where $a$ is an arbitrary constant vector.
From this identity and because $\nabla \times \mathbf{a = 0 }, $ thus $ \nabla\times F=0 + \nabla f \times a$.
Thus $(\nabla\times F)\cdot d\mathbf{ S }= (\nabla f \times a)  \cdot d\mathbf{ S }
= (d\mathbf{ S }\times  \nabla f)  \cdot a $, thanks to the answer below.
Then Stokes's theorem for arbitrary a implies
$
\int_{C}f \mathbf{ a } \; d\mathbf{ r } =  \iint_S   (d\mathbf{ S }\times  \nabla f)  \cdot a 
$.

My concern: How do I proceed from here? Please explain steps in detail?

 A: Remember that
$$
(v\times w)\cdot z = \det(v,w,z) = \det (z,v,w) = (z\times v)\cdot w
$$
and apply this to $v= \nabla f$, $w=a$, $z=ds$
A: Apply the Kelvin-Stokes theorem to the vector field $fa$:
$$\oint fa \cdot dr = \int (\nabla \times [af]) \cdot dS$$
Use vector calculus identities to write
$$\nabla \times [af] = f \nabla \times a + \nabla f \times a$$
Since $\nabla \times a = 0$, we get
$$\oint f a \cdot dr = \int (\nabla f \times a) \cdot dS$$
The triple product can be cyclically permuted, yielding
$$\oint fa \cdot dr = \int (dS \times \nabla f) \cdot a$$
Since $a$ is constant, it can be moved out of both integrals, the way you would move a constant scalar out of an integral:
$$a \cdot \oint f \, dr = a \cdot \int dS \times \nabla f$$
$a$ was chosen arbitrarily; this is true for all $a$ and thus we can "cancel" $a$.  If one must think of it more rigorously, look at this expression above as a linear function of $a$.  Take a gradient of that function with respect to $a$.  The resulting gradients are equal on both sides.  The result is
$$\oint f \, dr = \int dS \times \nabla f$$
