Question: Find convergence area and absolute convergence area of the following series:
$$\sum\_{k=0}^{\infty}\frac{(2x)^{k}}{(1+x^{2})^{k}}$$
I wrote it as 
$$\sum\_{k=0}^{\infty}\frac{(2x)^{k}}{(1+x^{2})^{k}}=\sum\_{k=0}^{\infty}\left(\frac{2x}{1+x^{2}}\right)^{k}$$
and now this is geometric series, where $q=\frac{2x}{1+x^{2}}$. So it converges absolutly when $\frac{2|x|}{1+x^{2}}<1$ and convergence area and absolute convergence area are both $\mathbb{R}\setminus\{-1, 1\}$.

Is my solution correct?

Answer: 
Yes, this is correct. I would include additional reasoning as to how you solved the inequality to get the area of absolute convergence; e.g., $2|x|/(1+x^2) < 1$ implies $$0 < x^2 - 2|x| + 1 = (|x| - 1)^2,$$ and since such a square is only $0$ when $|x| = 1$ or $x \in \{-1,1\}$, the result follows.
You need to solve the inequality 
$$
x^2 + 1 - 2 x >0
$$
which is true for all $x$. Now add the boundary inequalities that come from the equliaty
$$
\frac{2x}{x^2+1}=0
$$
for $x=1$ or $x=-1$