Find an inverse for a function defined by an integral

Define a function for $x \geq 0$ by

$f(x) = \int_0^x (1+t^3)^{-\frac{1}{2}} \, dt.$

Prove that $f(x)$ is strictly increasing on the nonnegative real axis.
If $g$ denotes the inverse of $f$ prove that $g''$ is proportional to $g^2$ and find the constant of proportionality.

Proof. To show $f(x)$ is strictly increasing we take the derivative,
$f(x) = \int_0^x \frac{1}{\sqrt{1+t^3}} \, dt \quad \implies \quad f'(x) = \frac{1}{\sqrt{1+x^3}}.$

Since $f'(x) > 0$ for all $x \geq 0$ we have that $f(x)$ is strictly increasing on the nonnegative real axis $. \qquad \blacksquare$
Proof. If $g$ is the inverse of $f$ then we know (Theorem 6.7 on page 252 of Apostol)
$g'(y) = \frac{1}{f'(x)}.$

But, we have defined $g$ to the function such that $g(f(x)) = x$ . Therefore,

$g'(y) = \frac{1}{f'(g(y))}.$

Therefore, we have that

$f'(x) = \frac{1}{\sqrt{1+x^3}} \quad \implies \quad g'(y) = \sqrt{1+g^3(y)}.$

Taking another derivative of $g$ with respect to $y$ we then have

$\begin{align*} g''(y) &= \frac{3 (g(y))^2 \cdot g'(y)}{2 \sqrt{1+(g(y))^3}} \\[9pt] &= \frac{3}{2} \cdot (g(y))^2 \cdot \frac{g'(y)}{\sqrt{1+(g(y))^3}} \\[9pt] &= \frac{3}{2} \cdot (g(y))^2 \cdot \frac{\sqrt{1+(g(y))^3}}{\sqrt{1+(g(y))^3}} \\[9pt] &= \frac{3}{2} (g(y))^2. \qquad \blacksquare \end{align*}$