Answer
$$\frac{{dy}}{{dx}} = \frac{{2{{\sec }^2}x}}{{\tan x}}$$
Work Step by Step
$$\eqalign{
& y = \ln \left| {{{\tan }^2}x} \right| \cr
& {\text{differentiate with respect to }}x \cr
& \frac{{dy}}{{dx}} = {D_x}\left( {\ln \left| {{{\tan }^2}x} \right|} \right) \cr
& {\text{use the chain rule }}{D_x}\left( {\ln u} \right) = \frac{1}{u} \cdot {D_x}\left( u \right).{\text{ consider }}u = \sin {x^2} \cr
& \frac{{dy}}{{dx}} = \left( {\frac{1}{{{{\tan }^2}x}}} \right){D_x}\left( {{{\tan }^2}x} \right) \cr
& {\text{use the general power rule for derivatives }}\frac{d}{{dx}}\left[ {{u^n}} \right] = n{u^{n - 1}}\frac{{du}}{{dx}}.{\text{ consider }}u = \tan x \cr
& \frac{{dy}}{{dx}} = \left( {\frac{1}{{{{\tan }^2}x}}} \right)\left( {2\tan x} \right){D_x}\left( {\tan x} \right) \cr
& {\text{then}} \cr
& \frac{{dy}}{{dx}} = \left( {\frac{1}{{{{\tan }^2}x}}} \right)\left( {2\tan x} \right)\left( {{{\sec }^2}x} \right) \cr
& {\text{simplifying}} \cr
& \frac{{dy}}{{dx}} = \frac{{2{{\sec }^2}x}}{{\tan x}} \cr} $$
