Answer
$$ f'(x)= 4(2 e^{3x}+2 e^{-2x})^3(6 e^{3x}-4e^{-2x}).$$
Work Step by Step
Recall that $(e^x)'=e^x$
Recall that $(x^n)'=nx^{n-1}$
Since we have
$$ f(x)=(2 e^{3x}+2 e^{-2x})^4$$
then the derivative $ f'(x)$, using the chain rule, is given by
$$ f'(x)=4(2 e^{3x}+2 e^{-2x})^3(2 e^{3x}+2 e^{-2x})'=4(2 e^{3x}+2 e^{-2x})^3(6 e^{3x}-4e^{-2x}).$$
