Answer
$$\frac{{dy}}{{dx}} = \frac{2}{{\left( {\ln 2} \right)x}}$$
Work Step by Step
$$\eqalign{
& y = {\log _2}\left( {{x^2}/2} \right) \cr
& {\text{Find the derivative of }}y{\text{ with respect to }}x \cr
& \frac{{dy}}{{dx}} = \frac{d}{{dx}}\left[ {{{\log }_2}\left( {{x^2}/2} \right)} \right] \cr
& {\text{we can use the formula }}\cr
&\frac{d}{{dx}}\left[ {{{\log }_a}u} \right] = \frac{1}{{\ln a}} \cdot \frac{1}{u}\frac{{du}}{{dx}}{\text{ }}\left( {page\,\,390} \right) \cr
& {\text{where }}u{\text{ is any differentiable function of }}x \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\ln 2}}\left( {\frac{2}{{{x^2}}}} \right)\frac{d}{{dx}}\left[ {\frac{{{x^2}}}{2}} \right] \cr
& {\text{then}} \cr
& \frac{{dy}}{{dx}} = \frac{1}{{\ln 2}}\left( {\frac{2}{{{x^2}}}} \right)\left( x \right) \cr
& \frac{{dy}}{{dx}} = \frac{2}{{\left( {\ln 2} \right)x}} \cr} $$
