<s>[INST] Derivative of $f(x)=-6\sin^4 x$ 
$f(x)=-6\sin^4x$  

$f(x)=-6\sin x^4$  
$f'(x)=-6\cos x^4(4x^3)=-24x^3\cos x^4$  
What am I doing wrong? Please show the steps.
 [/INST] Whether $f(x) = -6\sin^4 x = -6(\sin x)^4$ or $f(x) = -6 \sin (x^4)$, the chain rule is the way to go.
If
$f(x) = -6(\sin x)^4, \tag{1}$
set
$u = \sin x \tag{2}$
so that (1) becomes
$f(u(x)) = -6(u(x))^4; \tag{3}$
now we apply the chain rule, which says that $df(u(x))/dx$ is given by
$\dfrac{df(u(x))}{dx} = \dfrac{df}{du}(u(x)) \dfrac{du}{dx}(x). \tag{4}$
We see that
$\dfrac{df(u(x))}{du}(u(x)) = -24(u(x))^3 = -24 (\sin x)^3 \tag{5}$
and
$\dfrac{du}{dx} = \cos x, \tag{6}$
whence, by (4),
$\dfrac{df(u(x))}{dx} = -24 (\sin x)^3 \cos x. \tag{7}$
On the other hand if
$f(x) = -6\sin(x^4), \tag{8}$
then we take
$u(x) = x^4 \tag{9}$
so that (8) yields
$f(u(x)) = -6 \sin u(x), \tag{10}$
from which we have
$\dfrac{df(u(x))}{du}(u(x)) = -6 \cos u(x) = -6 \cos x^4; \tag{11}$
also,
$\dfrac{du}{dx} = 4x^3; \tag{12}$
pulling (11) and (12) together with the aid of (4) gives the result
$\dfrac{df((x))}{dx} = -24x^3 \cos x^4. \tag{13}$
Either way, The Chain Rule rules!  Long Live The Chain Rule! 
Hope this helps.  Cheers,
and as always,
Fiat Lux!!!
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