Show that $\displaystyle \nabla\times(f\mathbf{v})=\nabla f\times\mathbf{v}+f\nabla\times\mathbf{v}$
Here $\displaystyle f(x,y,z)$ is a scalar field and $\displaystyle \mathbf{v}(x,y,z)$ is a vector field.
Rewrite your expression in component form:
$\displaystyle
\nabla \times (f \vec v) = \epsilon_{ijk} \partial_j (f \vec v_k)
$
where $\displaystyle
\epsilon_{ijk}
$ is the fully antisymmetric symbol which is 0 if any indices are the same, 1 if ijk is cyclic to xyz and -1 if ijk is cyclic to xzy.
Concentrate on what the derivative is doing. This is just like d(uv) = (du)v + u(dv):
$\displaystyle
\partial_j (f \vec v_k) = (\partial_j f) \vec v_k + f (\partial_j \vec v_k)
$
So the proof goes:
$\displaystyle
\nabla \times (f \vec v) = \epsilon_{ijk} \partial_j (f \vec v_k)
$
$\displaystyle
= \epsilon_{ijk} ( (\partial_j f) \vec v_k + f (\partial_j \vec v_k)
)$
$\displaystyle
= \epsilon_{ijk} (\partial_j f) \vec v_k + \epsilon_{ijk} f (\partial_j \vec v_k)
$
$\displaystyle
= \epsilon_{ijk} (\partial_j f) \vec v_k + f \epsilon_{ijk} (\partial_j \vec v_k)
$
$\displaystyle
= (\nabla f) \times \vec v + f \nabla \times \vec v$