Answer
The answers:
$y=f(x)=x^2+x^{-1}$,
$dy/dx=2x-x^{-2}$.
Work Step by Step
First way:
Using Eq. (8) we have
$dy/dx=y'(s)/x'(s)=(6s^5 - 3s^{-4})/(3s^2)=2 s^3-s^{-6}$.
Second way:
Since $x=s^3$, so $s=x^{1/3}$.
Substituting $s$ into $y=s^6+s^{-3}$ gives
$y=(x^{1/3})^6+(x^{1/3})^{-3}=x^2+x^{-1}$,
$y=f(x)=x^2+x^{-1}$.
Differentiating $f(x)$ we get
$dy/dx=2x-x^{-2}$.
Since $x=s^3$, this is the same as $2 s^3-s^{-6}$.
