Calculating sum of converging series $\sum_{n=1}^{\infty}\frac{1-n}{9n^3-n}$

Let's rewrite $\frac{1}{n}$ as $\frac{3}{3n}$, so your sum is $S+2$ with$$S:=\sum_{n\ge 1}\left(\frac{1}{3n-1}-\frac{3}{3n}+\frac{2}{3n+1}\right)\\=\sum_{n\ge 1}\int_0^1 x^{3n-2}\left(1-3x+2x^2\right)\mathrm dx=\int_0^1\frac{x-3x^2+2x^3}{1-x^3}\mathrm dx.$$Thus $$S+2=\int_0^1\frac{2+3x}{1+x+x^2}\mathrm dx=\frac32\int_0^1\frac{1+2x}{1+x+x^2}\mathrm dx+\frac12\int_0^1\frac{\mathrm dx}{1+x+x^2}.$$The rest I'll leave to you.

For the direct evaluation of the limit, you have received the good solution from J.G.

You could also consider the partial sum using, as you did, partial fraction decomposition $$\frac{1-n}{9n^3-n}=\frac{1}{3 n-1}+\frac{2}{3 n+1}-\frac{1}{n}$$ which makes $$S_p=\sum_{n=1}^{p}\frac{1-n}{9n^3-n}=\frac{1}{3} \left(\psi \left(p+\frac{2}{3}\right)-\psi \left(\frac{2}{3}\right)\right)+\frac{2}{3} \left(\psi \left(p+\frac{4}{3}\right)-\psi \left(\frac{4}{3}\right)\right)-H_p$$ where appear the digamma function and harmonic number.

Now, using asymptotics and Taylor series you should arrive to $$S_p=-\left(\gamma +\frac{1}{3}\psi\left(\frac{2}{3}\right)+\frac{2}{3} \psi \left(\frac{4}{3}\right)\right)+\frac{1}{9 p}-\frac{1}{9 p^2}+O\left(\frac{1}{p^3}\right)$$ which shows the limit and how it is approached.

Now, the particular values $$\psi\left(\frac{2}{3}\right)=-\gamma +\frac{\pi }{2 \sqrt{3}}-\frac{3 \log (3)}{2}$$ $$\psi\left(\frac{4}{3}\right)=3-\gamma -\frac{\pi }{2 \sqrt{3}}-\frac{3 \log (3)}{2}$$ make $$S_p=\left(-2+\frac{\pi }{6 \sqrt{3}}+\frac{3 \log (3)}{2}\right)+\frac{1}{9 p}-\frac{1}{9 p^2}+O\left(\frac{1}{p^3}\right)$$

For $p=10$, the exact value is $-\frac{477820712081}{12033629407800}\approx -0.039707$ while the above truncated expansion gives $-\frac{199}{100}+\frac{\pi }{6 \sqrt{3}}+\frac{3 \log (3)}{2}\approx -0.039782$.