Do do I find Interval of convergence for this problem

I have f(x) = x^2/(1+x^3)

I got the summation of

sigma( n= 0 to infinity) (-1)^x (x^(3n+2))

how do I find the interval of convergence for that? I only learned that the ratio test only apply in situations where it's something times x^n only?

Your series representation is almost correct; it should be:
sum(n=0 to infinity) [(-1)^n * x^(3n + 2)].

With a_n = (-1)^n * x^(3n + 2), we see that:
a_n+1 = (-1)^(n + 1) * x^[3(n + 1) + 2] = (-1)^(n + 1) * x^(3n + 5).

By the Ratio Test:
lim (n-->infinity) (a_n+1)/(a_n)
= lim (n-->infinity) [(-1)^(n + 1) * x^(3n + 5)]/[(-1)^n * x^(3n + 3)]
= -x^3.

Thus, the series converges absolutely when:
|-x^3| < 1 ==> -1 < x < 1.

Checking the endpoints:
(a) x = -1 ==> sum(n=0 to infinity) [(-1)^n * (-1)^(3n + 2)], which diverges
(b) x = 1 ==> sum(n=0 to infinity) (-1)^n, which also diverges.

Therefore, the series converges for -1 < x < 1.

I hope this helps!