Lets see if I learned from the last one.

$\displaystyle 2x^2y=sin2x$

$\displaystyle \frac{d}{dx}[2x^2y] = \frac{d}{dx}[sin2x] $

$\displaystyle 4xy\frac{dy}{dx}[2x^2]=\frac{dy}{dx}[cos2x]\frac{dy}{dx}[sin2]$

$\displaystyle \frac{dy}{dx}(2x^2-cos2x-sin2) = -4xy$

$\displaystyle \frac{dy}{dx} = -\frac{4xy}{2x^2-cos2x-sin2}$