Answer
$\frac{d^2y}{dx^2} =\frac{2y}{(y-x)^2} - \frac{y^2}{(y-x)^3}$
Work Step by Step
Applying the rules of derivatives:
$2xy-y^2=3$
$\frac{d}{dx}[2xy-y^2] =\frac{d}{dx}[3] $
$2y + 2x\frac{dy}{dx}-2y\frac{dy}{dx} = 0$
$\frac{dy}{dx}(2x-2y) = -2y$
$\frac{dy}{dx} = -\frac{y}{x-y}= \frac{y}{y-x}$
$\frac{d^2y}{dx^2} = \frac{\frac{dy}{dx}(y-x)-(\frac{dy}{dx}-1)y}{(y-x)^2} = \frac{ \frac{y}{y-x}(y-x)-( \frac{y}{y-x}-1)y}{(y-x)^2} = \frac{y+y-\frac{y^2}{y-x}}{(y-x)^2} = \frac{2y}{(y-x)^2} - \frac{y^2}{(y-x)^3}$
