To find the coefficient of \(x\) in the expansion of \(\left( \frac{1}{x} + 3x^2 \right)^5\), we use the binomial theorem:

\(\sum_{k=0}^{5} \binom{5}{k} \left( \frac{1}{x} \right)^{5-k} (3x^2)^k\)

We need the term where the power of \(x\) is 1. The general term is:

\(\binom{5}{k} \left( \frac{1}{x} \right)^{5-k} (3x^2)^k = \binom{5}{k} \cdot \frac{1}{x^{5-k}} \cdot 3^k \cdot x^{2k}\)

This simplifies to:

\(\binom{5}{k} \cdot 3^k \cdot x^{2k - (5-k)} = \binom{5}{k} \cdot 3^k \cdot x^{3k - 5}\)

We set the exponent of \(x\) to 1:

\(3k - 5 = 1\)

Solving for \(k\):

\(3k = 6\)

\(k = 2\)

Substitute \(k = 2\) back into the expression for the coefficient:

\(\binom{5}{2} \cdot 3^2 = 10 \cdot 9 = 90\)

Thus, the coefficient of \(x\) is 90.