I was given the example as an illustration of structure of permutations in my lecture notes on algebra as shown below:
$\bigl(\begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\ 2 & 1 & 3 & 4 & 9 & 6 & 5 & 8 & 7 \end{smallmatrix}\bigr)$ $= (12)(597) = (12)(57)(59) = (12)(36)(68)(36)(38)(57)(59)$
I get the first relationship was obtained by cycle decomposition, but how were 2nd and 3rd relationships obtained? If I may use some hints
On
I will try giving as many details as possible.
Second equality is just the equality $(a_1,\dots,a_n) = (a_1,a_n)(a_1,a_{n-1})\dots(a_1,a_2)$ (which is easy to prove) applied to the second factor. There are many other equalities of the same kind, for example $(a_1,\dots,a_n) = (a_1,a_2)(a_2,a_3)\dots(a_{n-1},a_n)$
Third equality is just multiplying in-between by the 4-terms factor $(36)(68)(36)(38)$, which can be seen to be the identity permutation the following way : $(36)(68)(36) = (38)$ as a special case of $\sigma(a_1,\dots,a_n)\sigma^{-1} = (\sigma(a_1),\dots,\sigma(a_n))$
I doubt that there was any particular general method used. I should think the point is to show that a given permutation can be written as a product of cycles in many ways. To find an example you could take a given permutation $\pi$ and a random choice of cycles $\gamma_1,\ldots,\gamma_s$, write $$\pi=\gamma_1\cdots\gamma_s\sigma\ ,$$ then compute $\sigma$ (and write it as a product of cycles if you like).
The end of your example is possibly useful in that it shows how to write a $3$-cycle as a product of transpositions, $$(597)=(57)(59)\ ,$$ and the bit in the middle can be interpreted siuilarly, $$(36)(68)(36)(38)=(683)(386)=(683)(683)^{-1}=\iota\ .$$ But to sum up again, I think it's just a pretty random example.
BTW a cycle can be written as a product of transpositions in many ways, but IMHO the easiest way to remember is not the one used in your example but $$(a_1a_2a_3\cdots a_n)=(a_1a_2)(a_2a_3)(a_3a_4)\cdots(a_{n-1}a_n)\ .$$