Answer
$curl (fF)=f curl F+(\nabla f) \times F$
Work Step by Step
The vector field $F$ is conservative when $curl F=0$
When $F=ai+bj+ck$, then we have $curl F=[c_y-b_z]i+[a_z-c_z]j+[b_x-a_y]k$
Plug $F=F_1i+F_2j+F_3z$
$$curl (fF)=curl (f F_1i+f F_2 j+fF_3 k)$$
or, $=[\dfrac{\partial [fF_3]}{\partial y}-\dfrac{\partial [fF_2]}{\partial z}]i+[\dfrac{\partial [fF_3]}{\partial x}-\dfrac{\partial [fF_1]}{\partial z}]j+[\dfrac{\partial [fF_2]}{\partial x}-\dfrac{\partial [fF_1]}{\partial y}]k$
or, $=F_3[\dfrac{\partial f}{\partial y}i-\dfrac{\partial f}{\partial x}j]+F_1[\dfrac{\partial f}{\partial z}j-\dfrac{\partial f}{\partial y}k]+F_2[\dfrac{\partial f}{\partial x}k-\dfrac{\partial f}{\partial z}i]+f curl F$
or, $=f curl F+(\nabla f) \times F$
Hence, $curl (fF)=f curl F+(\nabla f) \times F$ (Proved)
