To find the coefficient of \(x^2\) in the expansion of \(\left( x + \frac{2}{x} \right)^6\), we use the binomial theorem.

The general term in the expansion is given by:

\(T_k = \binom{6}{k} \left(x\right)^{6-k} \left(\frac{2}{x}\right)^k\)

We need the term where the power of \(x\) is 2:

\((6-k) - k = 2\)

\(6 - 2k = 2\)

\(2k = 4\)

\(k = 2\)

Substitute \(k = 2\) into the general term:

\(T_2 = \binom{6}{2} \left(x\right)^{6-2} \left(\frac{2}{x}\right)^2\)

\(= \binom{6}{2} x^4 \cdot \frac{4}{x^2}\)

\(= \binom{6}{2} \cdot 4 \cdot x^2\)

\(= 15 \cdot 4 \cdot x^2\)

\(= 60x^2\)

Thus, the coefficient of \(x^2\) is 60.