Answer
$curl (fF)=f curl F+(\nabla f) \times F$
Work Step by Step
Need to prove that $curl (fF)=\nabla f \times F+f curl F$
Let us consider that $F=A i+B j+C k$
Suppose $F=F_1i+F_2j+F_3z$
$curl (fF)=curl (f F_1i+f F_2 j+fF_3 k)=[\dfrac{\partial [fF_3]}{\partial y}-\dfrac{\partial [fF_2]}{\partial z}]i+[\dfrac{\partial [fF_3]}{\partial x}-\dfrac{\partial [fF_1]}{\partial z}]j+[\dfrac{\partial [fF_2]}{\partial x}-\dfrac{\partial [fF_1]}{\partial y}]k$
$curl (fF)=F_3[\dfrac{\partial f}{\partial y}i-\dfrac{\partial f}{\partial x}j]+F_1[\dfrac{\partial f}{\partial z}j-\dfrac{\partial f}{\partial y}k]+F_2[\dfrac{\partial f}{\partial x}k-\dfrac{\partial f}{\partial z}i]+f curl F$
Hence, $curl (fF)=f curl F+(\nabla f) \times F$
