Consider the binomial expansion of \(\left( \frac{x}{a} + \frac{a}{x^2} \right)^7\).

The general term is given by:

\(T_k = \binom{7}{k} \left( \frac{x}{a} \right)^{7-k} \left( \frac{a}{x^2} \right)^k\)

Simplifying, we have:

\(T_k = \binom{7}{k} \frac{x^{7-k}}{a^{7-k}} \frac{a^k}{x^{2k}} = \binom{7}{k} \frac{a^{k-(7-k)}}{x^{2k-(7-k)}} = \binom{7}{k} \frac{a^{2k-7}}{x^{k+7-2k}}\)

For the coefficient of \(x^4\), set \(k+7-2k = 4\), giving \(k = 3\).

The coefficient of \(x^4\) is:

\(\binom{7}{3} \frac{a^{6-7}}{x^4} = \frac{35}{a}\)

For the coefficient of \(x\), set \(k+7-2k = 1\), giving \(k = 6\).

The coefficient of \(x\) is:

\(\binom{7}{6} \frac{a^{12-7}}{x} = 7a^5\)

Given:

\(\frac{\frac{35}{a}}{7a^5} = 3\)

Simplifying:

\(\frac{5}{a^6} = 3\)

\(a^6 = \frac{5}{3}\)

\(a^2 = \frac{1}{9}\)

\(a = \pm \frac{1}{3}\)