Answer: \(-16640\).

Use the binomial theorem carefully, keeping track of the power of each factor in the general term.

The general term in the expansion of

\(\left(2x-\frac1x\right)^{10}\)

is

\(\binom{10}{r}(2x)^{10-r}\left(-\frac1x\right)^r.\)

This simplifies to

\(\binom{10}{r}2^{10-r}(-1)^r x^{10-2r}.\)

Now

To get \(x^8\) from the first part, we need

\(10-2r=8,\)

so \(r=1\). The coefficient is

\(\binom{10}{1}2^9(-1)=-5120.\)

To get \(x^8\) from

\(-x^2\left(2x-\frac1x\right)^{10},\)

we need the coefficient of \(x^6\) in the expansion. This occurs when

\(10-2r=6,\)

so \(r=2\). The coefficient of \(x^6\) is

\(\binom{10}{2}2^8=45\cdot256=11520.\)

Because this term is multiplied by \(-x^2\), its contribution to the \(x^8\) coefficient is

\(-11520.\)

Therefore the coefficient of \(x^8\) is

\(-5120-11520=-16640.\)