Your instincts are correct that you want to look for sources of discontinuity: zero in the denominator, <0 under an even root, ≤0 in a logarithm, etc. Your answer would be correct for a square or quartic root.
Even roots and odd roots behave differently
Even roots (x1/2,x1/4,x1/6,...) cannot have negative inputs, another way to say this is their Domain is [0.∞)
This is because
(-a)2=a2
a2=a2
When you do the inverse you get the square root of a2=±a In order to be a function we cannot have to outputs for one input so we choose to define f(x)=√(x) as only the positive root
Odd roots (x1/3,x1/5,x1/7,...) don't have this limitation
(-a)3=-a3
a3=a3
When you do the inverse you get the cubed root of a3=a, and the cubed root of -a3=-a, because of this the domain and range of odd root functions are (-∞,∞).
So since there is no discontinuity, since odd root functions don't have any limitations on their domain, the function is continuous from (-∞,∞).
Be careful when using your graphing calculator to determine discontinuity. Although they are great at showing end behavior and asymptotic discontinuity, removable discontinuities can be hard to see sometimes, so you want to always first use your logic check (is there a denominator, even root, log, or anything else that might cause problems involved) so you know what you are looking for.
Ex: (x-5)/(x2-25) would show the asymptote at x=-5 with little issue, but the hole at x=5 might be harder to see, so you may say this is continuous from (-∞,-5)(-5,∞) when the correct answer would be continuous from (-∞,-5)(-5,5)(5,∞)
