It is given that $f(x) = \int_0^x \frac{1}{\sqrt{1+t^3}}dt$ and $g(x)$ is the inverse of $f(x)$, then what is the value of $\frac{(g(x))^2}{g''(x)}$?
$f'(x) = (1+x^3)^{-0.5}$
$∵ g(x) = f^{-1} (x)$
$⇒ g'(x) = \frac{1}{f'(x)} = \sqrt{1+x^3}$
$⇒ g''(x) = -\frac{3x^2}{2\sqrt{1+x^3}}$
I am unable to proceed further. How to find $g(x)$ ?