Answer
$$\operatorname{curl}(f \mathbf{F})=f \operatorname{curl}(\mathbf{F})+(\nabla f) \times \mathbf{F}$$
Work Step by Step
We verify as follows:
\begin{align*}
\operatorname{curl}(f \mathbf{F})&= \begin{array}{|||}
i&j &k\\
\frac{\partial}{\partial x}& \frac{\partial}{\partial y}& \frac{\partial}{\partial z}\\
fF_1& fF_2&fF_3
\end{array}
\\
&= \left( \frac{\partial( fF_3)}{\partial y}- \frac{\partial( fF_2)}{\partial z}\right)i - \left( \frac{\partial(fF_3 )}{\partial x}- \frac{\partial(fF_1)}{\partial z}\right)j + \left( \frac{\partial( fF_2 )}{\partial x}- \frac{\partial(fF_1 )}{\partial y}\right) k \\
&= \left( f\frac{\partial( F_3)}{\partial y}+F_3\frac{\partial( f)}{\partial y}- f\frac{\partial( F_2)}{\partial z}-F_2\frac{\partial( f)}{\partial z}\right)i - \left( f\frac{\partial(F_3 )}{\partial x}+F_3\frac{\partial(f )}{\partial x}- F_1\frac{\partial(f)}{\partial z}-f \frac{\partial(F_1)}{\partial z}\right)j + \left( f\frac{\partial( F_2 )}{\partial x}+F_2 \frac{\partial( f )}{\partial x}- f\frac{\partial(f)}{\partial y}- F_1\frac{\partial(f )}{\partial y}\right) k \\
&=\left[ \left( f\frac{\partial( F_3)}{\partial y} - f\frac{\partial( F_2)}{\partial z}- \right)i - \left( f\frac{\partial(F_3 )}{\partial x} -f \frac{\partial(F_1)}{\partial z}\right)j + \left( f\frac{\partial( F_2 )}{\partial x} - f\frac{\partial(f)}{\partial y}- \right) k \right] + \left[ \left( F_3\frac{\partial( f)}{\partial y} -F_2\frac{\partial( f)}{\partial z}\right)i - \left( F_3\frac{\partial(f )}{\partial x}- F_1\frac{\partial(f)}{\partial z} \right)j + \left( F_2 \frac{\partial( f )}{\partial x} - F_1\frac{\partial(f )}{\partial y}\right) k \right]\\
&=f\left[ \left( \frac{\partial( F_3)}{\partial y} - \frac{\partial( F_2)}{\partial z}- \right)i - \left( \frac{\partial(F_3 )}{\partial x} - \frac{\partial(F_1)}{\partial z}\right)j + \left( \frac{\partial( F_2 )}{\partial x} - \frac{\partial(f)}{\partial y}- \right) k \right] + \left[ \left( F_3\frac{\partial( f)}{\partial y} -F_2\frac{\partial( f)}{\partial z}\right)i - \left( F_3\frac{\partial(f )}{\partial x}- F_1\frac{\partial(f)}{\partial z} \right)j + \left( F_2 \frac{\partial( f )}{\partial x} - F_1\frac{\partial(f )}{\partial y}\right) k \right]\\
&=f \operatorname{curl}(\mathbf{F})+(\nabla f) \times \mathbf{F}
\end{align*}
