Answer
$2x\log_2(3-2x)-\frac{2x^2}{(\ln 2)(3-2x)}$
Work Step by Step
$\frac{d}{dx}x^2\log_2(3-2x)$
$=x^2\frac{d}{dx}\log_2(3-2x)+\log_2(3-2x)\frac{d}{dx}x^2$
$=x^2\frac{1}{(3-2x)\ln 2}\frac{d}{dx}(3-2x)+\log_2(3-2x)*2x$
$=x^2\frac{1}{(3-2x)\ln 2}*-2+\log_2(3-2x)*2x$
$=2x\log_2(3-2x)-\frac{2x^2}{(\ln 2)(3-2x)}$