How do you use the chain rule to differentiate $y = {\sin ^3}x + {\cos ^3}x$

Hint: This problem deals with differentiation of the given equation using chain rule. The chain rule tells us how to find the derivative of a composite function. It is applied on composite functions. That is if a function is a product of two or more functions, then the differentiation of the composite function is given by the chain rule of differentiation. The chain rule is given below:
$ \Rightarrow \dfrac{d}{{dx}}\left( {{f_1}(x).{f_2}(x)} \right) = {f_1}(x).\dfrac{d}{{dx}}\left( {{f_2}(x)} \right) + {f_2}(x).\dfrac{d}{{dx}}\left( {{f_1}(x)} \right)$

Complete step-by-step answer:
Given an equation of trigonometric function, which is $y = {\sin ^3}x + {\cos ^3}x$
Let $y = f(x)$
Let ${f_1}(x) = {\sin ^3}x$
Let ${f_2}(x) = {\cos ^3}x$
So here $f(x) = {f_1}(x) + {f_2}(x)$
We know that the differentiation of the above equation gives:
$ \Rightarrow \dfrac{d}{{dx}}f(x) = \dfrac{d}{{dx}}{f_1}(x) + \dfrac{d}{{dx}}{f_2}(x)$
Applying the same to the given equation, $f(x) = {\sin ^3}x + {\cos ^3}x$ as shown below:
$ \Rightarrow f(x) = {\sin ^3}x + {\cos ^3}x$
Now differentiate the above equation as shown below:
\[ \Rightarrow \dfrac{d}{{dx}}f(x) = \dfrac{d}{{dx}}\left( {{{\sin }^3}x} \right) + \dfrac{d}{{dx}}\left( {{{\cos }^3}x} \right)\]
Now while differentiating the functions ${f_1}(x)$ and ${f_2}(x)$, applying the chain rule here, as shown below:
\[ \Rightarrow f'(x) = 3{\sin ^2}x.\dfrac{d}{{dx}}\left( {\sin x} \right) + 3{\cos ^2}x.\dfrac{d}{{dx}}\left( {\cos x} \right)\]
Here while differentiating the function \[{\sin ^3}x\], not only applying the formula $\dfrac{d}{{dx}}\left( {{f^n}(x)} \right) = n{x^{n - 1}}f'(x)$, that is not only reducing the power of the function but also differentiating the function of $x$, as it is not in terms of $x$, but it is a function of $x$, hence differentiating the function also.
We know that differentiation of \[\sin x\] is \[\cos x\], whereas the differentiation of \[\cos x\] is \[ - \sin x\]. Applying these substitutions in the above equation, as shown below:
\[ \Rightarrow f'(x) = 3{\sin ^2}x.\left( {\cos x} \right) + 3{\cos ^2}x.\left( { - \sin x} \right)\]
\[ \Rightarrow f'(x) = 3\cos x{\sin ^2}x - 3\sin x{\cos ^2}x\]
Now take the term $3\cos x\sin x$ common , in the above equation, as shown below:

\[ \Rightarrow f'(x) = 3\cos x\sin x\left( {\sin x - \cos x} \right)\]

Note:
Please note that there are basic differentiation rules in chain rule of differentiation. The sum rule says the derivative of a sum of functions is the sum of their derivatives. The difference rule says the derivative of a difference of functions is the difference of their derivatives. Also remember the basic derivatives such as:
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sin x} \right) = \cos x$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\cos x} \right) = - \sin x$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\tan x} \right) = {\sec ^2}x$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\sec x} \right) = \sec x\tan x$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\cot } \right) = - \cos e{c^2}x$
$ \Rightarrow \dfrac{d}{{dx}}\left( {\cos ecx} \right) = - \cos ecx\cot x$