We have
[tex]\begin{gathered} f(x)=\sqrt[3]{4x} \\ g(x)=6x \end{gathered}[/tex]
so
[tex](\frac{f}{g})(x)=\frac{\sqrt[3]{4x}}{6x+1}[/tex]
the restriction is not divided between 0
6x+1=0
6x=-1
x=-1/6
so our restriction is
[tex]x\ne-\frac{1}{6}[/tex]
so the correct answer is D.
