Thoughts...

Let $S_n = \displaystyle \sum_{k=1}^n f(k)$ . We can calculate $f(n)$ using $f(n) = \dfrac {S_{n-1}}{n^2-1}$ . And we get $f(2) = 12$ , $f(3) = 6$ , $f(4)=3.6$ ... and that $\dfrac {f(1)}{f(1)} = 1$ , $\dfrac {f(1)}{f(2)} = 3$ , $\dfrac {f(1)}{f(3)} = 6$ , $\dfrac {f(1)}{f(4)} = 10$ ... implying that $\dfrac {f(1)}{f(n)} = T_n = \dfrac {n(n+1)}2$ , where $T_n$ is the $n$ th triangular number. Let us prove that the claim $f(n) = \dfrac {2f(1)}{n(n+1)}$ is true for all $n \ge 1$ .

$\begin{aligned} S_n & = \sum_{k=1}^n f(k) = \sum_{k=1}^n \frac {2f(1)}{k(k+1)} = 2f(1) \sum_{k=1}^n \left(\frac 1k - \frac 1{k+1}\right) \\ & = 2f(1) \left(1-\frac 1{n+1}\right) = \frac {2nf(1)}{n+1} = \frac {2n^2f(1)}{n(n+1)} = n^2 f(n) \end{aligned}$

Proving that $f(n) = \dfrac {2f(1)}{n(n+1)}$ satisfies the equation $S_n = n^2f(n)$ . Therefore, $f(8) = \dfrac {2(36)}{8(8+1)}= \boxed{1}$ .

Note that $(n+1)^2f(n+1) \; = \; \sum_{j=1}^{n+1}f(j) \; = \; \sum_{j=1}^n f(j) + f(n+1) \; = \; n^2f(n) + f(n+1)$ so that $(n+2)f(n+1) \; = \; nf(n)$ and so $(n+1)(n+2)f(n+1) \; = \; n(n+1)f(n) \hspace{2cm} n \ge 1$ It is now a simple induction to show that $f(n) = \tfrac{72}{n(n+1)}$ for all $n \ge 1$ , and hence that $f(8) = \boxed{1}$ .