Answer: \(10\).

First expand only the terms of \(\left(x+\frac{2}{x}\right)^5\) that can contribute to an \(x^2\) term after multiplication by \(x-\frac3x\).

Using the binomial theorem,

\(\left(x+\frac{2}{x}\right)^5 =x^5+10x^3+40x+80x^{-1}+80x^{-3}+32x^{-5}.\)

When multiplying by \(x\), the \(40x\) term gives

\(x\cdot40x=40x^2.\)

When multiplying by \(-\frac3x\), the \(10x^3\) term gives

\(-\frac3x\cdot10x^3=-30x^2.\)

So the coefficient of \(x^2\) is

\(40-30=10.\)