Find second derivative (d^2y/dx^2) of (-2x^2 + 4xy = 2x/y)

Answer 1

-2x^2 + 4xy = 2x/y
-4x + 4y + 4xy' = 2/y - 2x/y^2 y'
y' = (2xy^2-2y^3+y)/(2xy^2+x)
Now, using the quotient rule, we get
y" = [(2xy^2+4xyy'-6y^2y'+y')(2xy^2+x)-(2xy^2-2y^3+y)(2y^2+4xyy'+1)]/(2xy^2+x)^2

Substituting in for all the y' we get
y" = [(2xy^2+4xy((2xy^2-2y^3+y)/(2xy^2+x))-6y^2((2xy^2-2y^3+y)/(2xy^2+x))+((2xy^2-2y^3+y)/(2xy^2+x)))(2xy^2+x)-(2xy^2-2y^3+y)(2y^2+4xy((2xy^2-2y^3+y)/(2xy^2+x))+1)]/(2xy^2+x)^2

There's no elegant way to simplify this, but one possible result is

2y^2/[x^2(2y^2+1)^3] * [(4y^4+4y^2+4y+1)x^2 - 2(4y^4-8y^2-1)x + 2y(4y^4+4y^2-3)]