Calculus of a Single Variable, Chapter 2, 2.5, Section 2.5, Problem 48

xy-1=2x+y^2
Differentiating both sides with respect to x,
xy'+y=2+2y*y'
(x-2y)y'=2-y
y'=(2-y)/(x-2y)
Differentiating again with respect to x,
(d^2y)/dx^2=((x-2y)d/dx(2-y)-(2-y)d/dx(x-2y))/(x-2y)^2
(d^2y)/dx^2=((x-2y)(-y')-(2-y)(1-2y'))/(x-2y)^2
(d^2y)/dx^2=(-xy'+2yy'-2+4y'+y-2yy')/(x-2y)^2
(d^2y)/dx^2=((4-x)y'+y-2)/(x-2y)^2
Now plug in the value of y' in second derivative
(d^2y)/dx^2=(((4-x)(2-y))/(x-2y)+y-2)/(x-2y)^2
(d^2y)/dx^2=((4-x)(2-y)+(x-2y)(y-2))/(x-2y)^3
(d^2y)/dx^2=(8-4y-2x+xy+xy-2x-2y^2+4y)/(x-2y)^3
(d^2y)/dx^2=(8-4x+2xy-2y^2)/(x-2y)^3