Answer
$\frac{d^2y}{dx^2}=\frac{2(y-2)(x-y-2)}{(x-2y)^3}$
Work Step by Step
$xy-1=2x+y^2$
$\frac{d}{dx}[xy-1]=\frac{d}{dx}[2x+y^2]$
$y+x\frac{dy}{dx}=2+2y\frac{dy}{dx}$
$(x-2y)\frac{dy}{dx}=2-y$
$\frac{dy}{dx}=\frac{2-y}{x-2y}$
$\frac{d^2y}{dx^2}=\frac{d}{dx}[\frac{dy}{dx}]$
$\frac{d^2y}{dx^2}=\frac{d}{dx}[\frac{2-y}{x-2y}]$
$\frac{d^2y}{dx^2}=\frac{-\frac{dy}{dx}(x-2y)-(2-y)(1-2\frac{dy}{dx})}{(x-2y)^2}$
$\frac{d^2y}{dx^2}=\frac{-(\frac{2-y}{x-2y})(x-2y)-(2-y)(1-2(\frac{2-y}{x-2y}))}{(x-2y)^2}$
$\frac{d^2y}{dx^2}=\frac{-(2-y)-(2-y)(1-2(\frac{2-y}{x-2y}))}{(x-2y)^2}$
$\frac{d^2y}{dx^2}=\frac{-(2-y)-(2-y)(1-\frac{4-2y}{x-2y})}{(x-2y)^2}$
$\frac{d^2y}{dx^2}=\frac{y-2+(y-2)(1-\frac{4-2y}{x-2y})}{(x-2y)^2}$
$\frac{d^2y}{dx^2}=\frac{(y-2)(1+1-\frac{4-2y}{x-2y})}{(x-2y)^2}$
$\frac{d^2y}{dx^2}=\frac{(y-2)(2-2\frac{2-y}{x-2y})}{(x-2y)^2}$
$\frac{d^2y}{dx^2}=\frac{2(y-2)(1-\frac{2-y}{x-2y})}{(x-2y)^2}$
$\frac{d^2y}{dx^2}=\frac{2(y-2)((x-2y)-(2-y))}{(x-2y)^3}$
$\frac{d^2y}{dx^2}=\frac{2(y-2)((x-2y)+y-2)}{(x-2y)^3}$
$\frac{d^2y}{dx^2}=\frac{2(y-2)(x-y-2)}{(x-2y)^3}$
