Like already mentioned $f$ is not injective. For every $y>0$ there are exactly two $x\in \mathbb R$ with $y=f(x)$, one of the $x$ is greater than zero and the other one is less than zero. Let's take only the positive one.
For every $x>0$ holds $x^4<x^4+\frac 13 x^3+x^2$. For $y:=x^4+\frac 13 x^3+x^2$ we get $\sqrt [4]y<x$. By inserting $f^{-1}(y)>\sqrt [4]y$ we have
$$\int_{2a}^{8a}\frac{1}{(f^{-1}(y))^2+(f^{-1}(y))^4}\,\mathrm dy\leq \int_{2a}^{8a} \frac{1}{\sqrt y+y}\,\mathrm dy=\left[2 \ln(1 + \sqrt y)\right]_{2a}^{8a}=2\ln\left(\frac{1+2\sqrt{2a}}{1+\sqrt{2a}}\right)\xrightarrow{a\to \infty}2\ln(2)$$
On the other hand for every $C\in (0,1)$ and every sufficently large $x$, we have $y>Cx^4$. Hence,
$$\int_{2a}^{8a}\frac{1}{(f^{-1}(y))^2+(f^{-1}(y))^4}\,\mathrm dy\geq \int_{2a}^{8a} \frac{1}{C^2\sqrt y+C^4y}\,\mathrm dy= \left[\frac{2}{C^4} \ln(1 + C^2 \sqrt{y})\right]_{2a}^{8a}=\frac{2}{C^4}\ln\left(\frac{1+2C^2\sqrt{2a}}{1+C^²\sqrt{2a}}\right)\xrightarrow{a\to \infty}\frac{2}{C^4}\ln(2).$$
Since this holds for all $C<1$, we get $n=2^2=4$.
