I have some trouble showing at which $x \in \mathbb{R}$ the series $\sum_{n=1}^{\infty} \frac{ x^{2n}}{1+ x^{2n}}$ converges and why the resulting function is continuous on this set.
I think it is quite clear that for $x > 1$ we have $\lim_{n\rightarrow\infty}\frac{ x^{2n}}{1+ x^{2n}} = 1$ and for $ x = 1, \lim_{n\rightarrow\infty}\frac{ x^{2n}}{1+ x^{2n}} = 1/2$ thus in both cases the series can not converge.
Now for any $[-c, c]$ with $c \in (0,1)$ we know that for any $n \in \mathbb{N}:$
$$M_n :=\sup\{ \frac{ x^{2n}}{1+ x^{2n}} | x \in [-c, c]\} = \frac{ c^{2n}}{1+ c^{2n}}.$$
Then we have by a argument that is similar to proving convergence of the geometric series that: $$\sum_{n=1}^{\infty} M_n=\sum_{n=1}^{\infty} \frac{ c^{2n}}{1+ c^{2n}} < \sum_{n=1}^{\infty} c^{2n} = \frac{ c^{2}}{1 - c^{2}}$$ Thus over this closed interval $[-c, c]$ Weierstrass M-test tells us that we have uniform convergence. Now obviously for any $n \in \mathbb{N},$ $\frac{ x^{2n}}{1+ x^{2n}}$ is continous on all of $\mathbb{R}$ thus: $$f:= [-c, c] \rightarrow \mathbb{R}: x \rightarrow \sum_{n=1}^{\infty} \frac{ x^{2n}}{1+ x^{2n}}$$ is well defined and continuous.
However I am not sure if I can conclude from this that just because this holds for any $[-c, c]$ with $c \in (0,1)$ we automatically get uniform convergence and therefore continuity on $(-1,1)$. I think I can not use the M-test on this open interval directly as for all $n \in \mathbb{N}$: $$M_n :=\sup\{ \frac{ x^{2n}}{1+ x^{2n}} | x \in (-1, 1)\} = \frac{ 1}{2}.$$
Plotting the function seems to imply that it does indeed hold on $(0,1)$:
If the series converges uniformly on $(-1,1)$ then $\frac {x^{2n}} {1+x^{2n}} \to 0$ uniformly. From this you can conclude that $\frac {(1-1/n)^{2n}} {1+(1-1/n)^{2n}} \to 0$. However the limit is $\frac {e^{-2}} {1+e^{-2}}$. This contradiction shows that the series does not converge uniformly on $(-1,1)$.