I assume that $\sum_{k=0}^{\infty}\left(\frac{x^{2k}}{2^{2k}}-\frac{x^{2k+1}}{3^{2k+1}}\right)$ is intended. You can avoid worrying about uniform convergence by using the ratio test. Calling the $k$-th term $a_k$, and letting $u=x/2$, we have
$a_k=\left(\frac{x}2\right)^{2k}-\left(\frac{x}3\right)^{2k+1}=u^{2k}-\left(\frac23\right)^{2k+1}u^{2k+1}=u^{2k}\left(1-\left(\frac23\right)^{2k+1}u\right)\;,$
so
$\lim_{k\to\infty}\left|\frac{a_{k+1}}{a_k}\right|=\lim_{k\to\infty}u^2\left|\frac{1-(2/3)^{2k+3}u}{1-(2/3)^{2k+1}u}\right|=u^2\;.$
Clearly you need $u^2<1$, or $|x|<2$, and it’s not hard to check that you don’t get convergence at either endpoint.