Yhyjyk

This power series $$f(x)=sum_n=1^infty fracx^3n3n$$ when differentiated, loses $3n$ in the denominator, with one manipulation, one can get $$f'(x)=frac1x(1-x^3) $$ using the geometric series sum formula. Since this is $f'(x)$, integration is required so we get $$f(x)=log(x)-frac13log(1-x^3) + C$$
(the log base is probably $e$). The usual step for finding the constant is to find it's value for $x=0$ in the given power series and the new function. This obviously won't work here or in any other case where a function isn't defined for $x=0$. How do I find the constant then?

Differentiating $f(x)$:

$$f(x)=fracx^33+fracx^66+cdotsimplies f'(x)=x^2+x^5+cdots=sum_n=1^inftyx^3n-1=fracx^21-x^3$$ for $|x|<1$, which results in $$f(x)=-frac13ln(1-x^3)+C$$ after integration. Your function now is well-defined at $x=0$, and this implies $C=0$.

You're not quite on track. One way to know that for sure is that your function isn't defined at $x=0,$ but the power series certainly is!

For $|t|<1,$ we have that $$sum_n=0^infty t^n=frac11-t,$$ so $$sum_n=1^infty t^n=frac11-t-1=fract1-t.$$ Thus, we have for $|x|<1$ that $$f'(x)=sum_n=1^infty x^3n-1=frac1xsum_n=1left(x^3right)^n=frac1xcdotfracx^31-x^3=left(1-x^3right)^-1cdot-frac13fracdleft(1-x^3right)dx.$$ This has the antiderivative family $$f(x)=-frac13lnleft(1-x^3right)+C,$$ which is defined at every $x$ in the radius of convergence, allowing you to find your constant in the usual way.

More generally, let's suppose that you've used a power series $$sum_n=k^infty a_n(x-a)^n$$ for some real $a$ and some integer $kge 0,$ and have (by differentiation and then integration) determined that $$f(x)+C=sum_n=k^infty a_n(x-a)^n$$ for some constant $C.$ Even without determining the radius of convergence, if you haven't made any errors, then the function is necessarily defined at $x=a,$ so you can use $x=a$ to determine what your value of $C$ will be.

Namely, we will have $C=-f(a)$ if $kge 1,$ and $C=a_0-f(a)$ otherwise.