First consider the subsituted eqatuion, with $z^2=x$. I also replaced $j$ by $n$:
$$\sum_{n = 0}^\infty\frac{2^n}{3^n+4^n}(x)^n$$
The ratio test:
$$R_x=\lim_{n\to \infty}|a_n/a_{n+1}|=0.5\lim_{n\to \infty}\frac{3^{n+1}+4^{n+1}}{3^{n}+4^n}$$.
Now divide the expand the fraction with $4^{-n-1}$:
$$R_x=0.5\lim_{n\to \infty}\frac{(0.75)^{n+1}+1}{1/3(0.75)^{n+1}+0.25^\cdot }=2$$.
In the previous limit all terms $(0.75)^n \to 0$ as $n\to \infty$. We are left with $1/0.25=4$.
So the radius of convergence for x is $R_x=2$. Take the squareroot to get the radius of convergence for $z$ to be $R_z=\sqrt{2}$.