How do we find the inverse of $f(x)=x^3+x$?
$$x=f^{-1}(y) \implies y=(f^{-1}(y))^3+f^{-1}(y)$$
$$(f^{-1}(y))^3+f^{-1}(y)-y=0\tag{1}$$
At this point we need to find a $f^{-1}(y)$ that solves this equation. But how?
To provide some context about why I am asking:
if $g(x)=x^3$ and $f(x)=\sqrt[3]{x}$ then $g$ and $f \circ g$ are differentiable, and $f$ is not.
In order for $f$ to be differentiable we need to impose an extra condition on $g$, namely that $g' \neq 0$.
$g_1(x)=x^3+x$ is such that $g_1' \neq 0$.
At this point I wanted to obtain an analytical expression for $g_1^{-1}$, to show that it is indeed differentiable.
Since you have only one real root for the cubic
$$x^3+x-y=0$$ use the hyperbolic method to obtain $$x=\frac{2}{\sqrt{3}}\sinh \left(\frac{1}{3} \sinh ^{-1}\left(\frac{3 \sqrt{3} }{2}y\right)\right)$$ Just follow the steps.