Tried to apply the ratio test to determine the convergence interval, but get the limit as a constant. So here is the series: $\Sigma_{n=1}^{\infty} \frac{x^{2n}}{1+x^{4n}}$
$$\left| \frac{x^{2(n+1)} (1+x^{4n})}{x^{2n}(1+x^{4(n+1)})}\right| = \left|\frac{x(1+x^{4n})}{1+x^{4n+4}} \right| \overset{\text{ n } \rightarrow \infty }{\rightarrow}\left[\frac{\infty}{\infty} \right]= \frac{x+x^{4n+1}}{1+x^{4n+4}} = \lim_{n \to \infty} \frac{x}{1+x^{4n+4}} +\lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}} = \\ 0 + \lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}} = \frac{\infty}{\infty} =\left[\text{ apply L'Hospital rule (4n+1)th times } \right] = \frac{1}{(4n+4)(4n+3)(4n+2)x^3} = 0 $$
Where is my mistake? Maybe some other technique would be useful for these series?
$$\lim_{n \to \infty} \frac{x}{1+x^{4n+4}} +\lim_{ n\to \infty} \frac{x^{4n+1}}{1+x^{4n+4}}  \\ \text{ if } x = 0 \text{ then limit equals } 0 \\ \text{if x=1 then limit equals to 1 } \\ \text{if } 0<x<1 \text{ then limit is equal to x [ the same goes for } -1<x<0 \\ \text{ if } x> 1 \text{ then limit is equal to} \frac{1}{x^3} \text{ the same goes for x<1} $$

Let's proceed from your work.  Corrections are highlighted in red.  We have
$$h(x) = \left| \frac{x^{2(n+1)} (1+x^{4n})}{x^{2n}(1+x^{4(n+1)})}\right| = \left|\frac{x^{\color{red}{2}}(1+x^{4n})}{1+x^{4n+4}} \right|.$$  At this point, we need to consider separate cases.  The reason is becuase for $n \ge 1$, the behavior of $|x^n|$ depends on whether $|x| < 1$, $|x| = 1$, or $|x| > 1$.
Case $|x| = 1$:  Then $x^2 = x^4 = 1$ and we have $$h(x) = \frac{1(1+1)}{(1+1)} = 1.$$  Going back to the original sum, $$f_n(x) = \frac{x^{2n}}{1+x^{4n}}$$ implies $$f_n(-1) = f_n(1) = \frac{1}{2} > 0$$ so the sum diverges.
Case $|x| > 1$:  Then $$\lim_{n \to \infty} 1/x^n = 0$$ and we have, after dividing numerator and denominator by $x^{4n+4}$, $$h(x) = \left|\frac{1/x^{4n+2} + 1/x^2}{1/x^{4n+4} + 1}\right|.$$  Hence $$\lim_{n \to \infty} h(x) = \left|\frac{0 + 1/x^2}{0 + 1} \right| = \frac{1}{x^2} < 1$$ and $\sum f_n(x)$ converges.
Case $|x| < 1$:  Then $x^n \to 0$ as $n \to \infty$, hence $$\lim_{n \to \infty} h(x) = \left|\frac{x^2(1 + 0)}{1 + 0}\right| = x^2 < 1$$ and again the sum converges.
Of course, this is not the simplest nor most elegant solution, but it is conforming to how one would apply the ratio test to determine the convergence of the sum.
