if

$f(x) = \dfrac{g(x)}{h(x)}$

and

$f^\prime(x)=\dfrac{df}{dx}(x)$

then

$f^\prime(x)=\dfrac{g^\prime(x)h(x)-g(x)h^\prime(x)}{(h(x))^2}$

so in this case

$g(x)=x^3$

$g^\prime(x)=3x^2$

$h(x)=(x^2+4)^2$

$h^\prime(x)=2(x^2+4)2x$

you should be able to put all the pieces together from here.