Perhaps apply sum of perfect powers?

Taking the reciprocal of the first line yields $\dfrac{a^4-a^2+1}{a^2} = a^2 - 1 + \dfrac{1}{a^2} = \frac{37}{4}$ . Adding three to both sides and completing the square yields $\left (a+ \dfrac{1}{a} \right )^2 = \dfrac{49}{4}$ , so $a+\dfrac{1}{a} = \dfrac{7}{2}$ .

Cubing this expression gives us $a^3+3a+\dfrac{3}{a} + \dfrac{1}{a^3} = \dfrac{343}{8}$ . From before, we calculate that $3a + \dfrac{3}{a} = \dfrac{84}{8}$ , so $a^3+\frac{1}{a^3} = \frac{259}{8}$ . Subtracting 1 from both sides and reciprocating yields $\dfrac{a^3}{a^6-a^3+1} = \dfrac{8}{251}$ , so $m=8, n=251$ . Thus, $m+n=259$ .