Compute the given limit as $x$ approaches $\infty$ 
If $f(x) = 8x^3+3x$ then, $$\lim_{x \to \infty} \frac{f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}$$ is?

My attempt:
It is clear that the function cannot be easily inverted. So, there must be something in the limit given itself that may simplify the problem.
Honestly, I have no clue what to do here.
There are a few things which I could see is that the function has only $1$ root(i.e $0$) and is bijective on $x \in \mathbb R$. But that gave no benefit except showing that the inverse of the function exists.
Any help would be appreciated.

Consider the equation
$$y=8x^3+3x$$ As you say, there is only one real root which is given by
$$x(y)=\frac{1}{2} \left(\frac{\sqrt[3]{\sqrt{2} \sqrt{2 y^2+1}+2
   y}}{2^{2/3}}-\frac{1}{\sqrt[3]{2} \sqrt[3]{\sqrt{2} \sqrt{2 y^2+1}+2 y}}\right)$$
Now, tedious but doable,for each piece, use Taylor expansion for infinitely large values of $y$ to get
$$x(8y)=\sqrt[3]{y}-\frac{\sqrt[3]{\frac{1}{y}}}{8}+O\left(\frac{1}{y^{4/3}}\right)  $$
$$x(y)=\frac{\sqrt[3]{y}}{2}-\frac{\sqrt[3]{\frac{1}{y}}}{4}+O\left(\frac{1}{y^{4/3}}\right)$$
$$x(8y)-x(y)=\frac{\sqrt[3]{y}}{2}+\frac{\sqrt[3]{\frac{1}{y}}}{8}+O\left(\frac{1}{y^{4/3}}\right)$$ will show  the limit and also how it is approached.
Edit
May be tricky but without Taylor series for the root of the cubic equation. Let us consider that, for large values of $x$
$$8x^3+3x=8 x^3+3 x+O\left(\frac{1}{x}\right)$$ Now, series reversion gives
$$x(y)=\frac{\sqrt[3]{y}}{2}-\frac{\sqrt[3]{\frac{1}{y}}}{4}+O\left(\frac{1}{y}\right)$$ and continue for the same result.
