$$\sum_{n\ge 1}\frac{(n!)^3}{(3n)!}z^{3n}=\sum_{k=3n, n\ge 1}\frac{((k/3)!)^3}{k!}z^k.$$
The ratio test seems like a better option than the root test here, but $\frac{a_{k+1}}{a_{k}}$ is either $0$ or undefined. So the radius (the limsup) is taken to be $\infty$. Am I using the ratio test correctly or should I be using it with $\frac{a_{k+3}}{a_{k}}$ instead?
Note that, if $z\neq0$,\begin{align}\lim_{n\to\infty}\frac{\left|\frac{((n+1)!)^3}{(3(n+1))!}z^{3(n+1)}\right|}{\left|\frac{(n!)^3}{(3n)!}z^{3n}\right|}&=\lim_{n\to\infty}\frac{(n+1)^3}{(3n+3)(3n+2)(3n+1)}|z|^3\\&=\frac{|z|^3}{3^3}\\&=\left(\frac{|z|}3\right)^3.\end{align}Therefore, the radius of convergence is $3$.