Exploit the innate symmetry! This is solvable purely mechanically, i.e. no lucky guesses are needed. The rational functions are all symmetric functions of $\rm\:a,b,c.\:$ By the Fundamental Theorem of Symmetric Polynomials, every symmetric polynomial is a polynomial in the elementary symmetric polynomials $\rm\:e_i.\:$ There is a simple effective algorithm due to Gauss to compute this representation. Putting all over a common denominator, we are given
$\rm e_1\ =\ a + b + c$
$\rm \frac{e_2}{e_3}\ =\ \frac{ab + bc + ca}{abc}\ =\ c^{-1} + a^{-1} + b^{-1}$
and we seek
$\rm \frac{f(a,b,c)}{e_3}\ =\ \frac{a^2 (b+c) + a\: (b^2+c^2) + b^2 c + b c^2 }{abc}$
Now use Gauss's algorithm to rewrite $\rm\:f\:$ in terms of the elementary symmetric polynomials $\rm\:e_i.\:$ This works as follows: if $\rm\ a^i\ b^j\ c^k\ $ is the highest term w.r.t. lex order $\rm\ a > b > c\ $ then kill the highest term of $\rm\:f\:$ by subtracting $\rm\ e_1^{i-j}\ e_2^{j-k}\ e_3^k\:.\:$ Here, since $\rm\ a^2 b^1 c^0\: $ is the highest term in $\rm\:f,\:$ we subtract from $\rm\:f\:$ the term $\rm e_1^{2-1}\ e_2^{1-0}\ e_3^0\ = (a+b+c) (ab+bc+ca),\:$ leaving $\rm\: -3abc = -3\:e_3.\:$ Thus we infer$\rm\:f = e_1 e_2 - 3\:e_3,\:$ so $\rm\:f/e_3 = e_1 (e_2/e_3) - 3\: =\: 20\cdot 30 - 3\: =\: 597.$
While this problem is so simple that one may be lucky enough to stumble upon this representation, one will rarely be so lucky with problems of even slightly greater complexity.