For a brief understanding, let's just do a brief change of variables. the original partial fraction decomposition of your problem is
$$\frac{1}{n(n+3)(n+6)}=\frac{1}{18}\left[\frac{1}{n}-\frac{2}{n+3}+\frac{1}{n+6}\right]$$
Letting $n=3m$, we get
$$\frac{1}{n(n+3)(n+6)}=\frac{1}{54}\left[\frac{1}{n}-\frac{2}{n+1}+\frac{1}{n+2}\right]$$
Showing the sum as an array after factoring out the $1/54$,
$$
\begin{matrix}
n=1: & 1/1 & -2/2 & \color\red{1/3} \\
n=2: & 1/2 & \color\red{-2/3} & \color\orange{1/4} \\
n=3: & \color\red{1/3} & \color\orange{-2/4} & \color\green{1/5} \\
n=4: & \color\orange{1/4} & \color\green{-2/5} & \color{blue}{1/6} \\
n=5: & \color\green{1/5} & \color{blue}{-2/6} & \vdots \\
n=6: & \color{blue}{1/6} & \vdots & \vdots
\end{matrix}
$$
You can see then by summing, the diagonals add to 0 and so all terms below the red diagonal will cancel by summation to 0. the slight difference with your sum is that your fraction decomposition shows that the cancellation occurs between terms separated by 3 units; your array would look like this after factoring out the $1/18$:
$$
\begin{matrix}
n=1: & 1/1 & -2/4 & \color\red{1/7} \\
n=2: & 1/2 & -2/5 & \color\orange{1/8} \\
n=3: & 1/3 & -2/6 & \color{green}{1/9} \\
n=4: & 1/4 & \color\red{-2/7} & \color{blue}{1/10} \\
n=5: & 1/5 & \color\orange{-2/8} & \color{teal}{1/11} \\
n=6: & 1/6 & \color{green}{-2/9} & \color{aqua}{1/12} \\
n=7: & \color\red{1/7} & \color{blue}{-2/10} & \vdots \\
n=8: & \color\orange{1/8} & \color{teal}{-2/11} & \vdots \\
n=9: & \color{green}{1/9} & \color{aqua}{-2/12} & \vdots \\
n=10: & \color{blue}{1/10} & \vdots & \vdots \\
n=11: & \color{teal}{1/11} & \vdots & \vdots \\
n=10: & \color{aqua}{1/112} & \vdots & \vdots \\
\end{matrix}
$$
Thus, the sum of your series is the remaining black terms at the top as every skewed diagonal in the picture sums to 0.
I will note, Lab's answer is the best here as it really simplifies the problem mathematically and once $n=3$, you can see the cancellation occur nicely...
EDIT:
There is a nice way to look at a general case as well. Let
$$\prod_{j=0}^m \frac{1}{n+jk}$$
has its partial fraction decomposition as
$$\frac{1}{m!k^m}\left[\sum_{j=0}^m{(-1)^j\binom{m}{j}\frac{1}{n+jk}}\right]$$
The problem here is that we want to sum over $n$. So
$$S_{m,k}=\sum_{n=1}^\infty{\prod_{j=0}^m \frac{1}{n+jk}}={\frac{1}{m!k^m}\left[\sum_{n=1}^\infty\sum_{j=0}^m{(-1)^j\binom{m}{j}\frac{1}{n+jk}}\right]}$$
But now we can take advantage of the telescoping property of the sum. This then reduces from infinite sums to finite sums. We have for the bracketed double sum from above...
$$\sum_{n=1}^\infty\sum_{j=0}^m{(-1)^j\binom{m}{j}\frac{1}{n+jk}}=\sum_{i=1}^m\sum_{n=1}^{ik}\binom{m}{i}\frac{(-1)^{i-1}}{n}=\sum_{i=1}^m{(-1)^{i-1}}\binom{m}{i}\sum_{n=1}^{ik}\frac{1}{n}$$
Now using the $k$-th harmonic number formula,
$$H_k=\sum_{n=1}^k{\frac{1}{n}}$$
we see that
$$S_{m,k}=\frac{1}{m!k^m}\sum_{i=1}^m{(-1)^{i-1}}\binom{m}{i}H_{ik}$$
This agrees with the additional answer provided by since this case we have $m=2, k=3$, so
$$S_{2,3}=\frac{1}{2!3^2}\sum_{i=1}^2{(-1)^{i-1}}\binom{2}{i}H_{3i}=\frac{1}{18}\left(2H_3-H_6\right)=\frac{1}{18}\left(2\cdot\frac{11}{6}-\frac{49}{20}\right)=\frac{73}{1080}$$
