Verifying $ \sum_{k=0}^{\infty}{3k \choose k}\frac{9k^2-3k-1}{(3k-1)(3k-2)}\left(\frac{2}{27}\right)^k=\frac{1}{4}$ $$
\sum_{k=0}^{\infty}{3k \choose k}\dfrac{9k^2-3k-1}{(3k-1)(3k-2)}\left(\dfrac{2}{27}\right)^k=\dfrac{1}{4}
$$
After some simplification, I got the following result:
$$
\sum_{k=0}^{\infty}\left\{{3k \choose k}+\dfrac{27}{2k}{3k-4 \choose 2k-3}\right\}\left(\dfrac{2}{27}\right)^k.
$$ 
Now, how I can proceed?

By Lagrange's inversion theorem (see Bring radical for the quintic analogue)
$$ \sum_{k\geq 0}\binom{3k}{k}\left(\frac{4}{27}\right)^k x^{2k} = \frac{\cos\left(\frac{1}{3}\arcsin x\right)}{\sqrt{1-x^2}}  \tag{1}$$
and by partial fraction decomposition
$$ \frac{9k^2-3k-1}{(3k-1)(3k-2)} = 1+\frac{1}{3k-1}+\frac{1}{3k-2}\tag{2} $$
hence by evaluating $(1)$ at $x=\frac{1}{\sqrt{2}}$ we immediately get
$$ \sum_{k\geq 0}\binom{3k}{k}\left(\frac{2}{27}\right)^k = \frac{1+\sqrt{3}}{2}.  \tag{3}$$
By reindexing the original series equals
$$-\frac{1}{2}+\sum_{k\geq 0}\binom{3k}{k}\frac{3(3k+2)(3k+1)}{(2k+2)(2k+1)}\left[1+\frac{1}{3k+1}+\frac{1}{3k+2}\right]\left(\frac{2}{27}\right)^k \tag{4}$$
hence it is enough to compute the integrals over $(0,1/\sqrt{2})$ of the RHS of $(1)$ and of the RHS of $(1)$ multiplied by $x$. They are $\frac{3}{2\sqrt{2}}(\sqrt{3}-1)$ and $\frac{3}{16}(5-2\sqrt{3})$ by the substitution $x\mapsto\sin\theta$. By re-combining these pieces the claim is proved.
