# 2xy+2y^2=13 #

Differentiating wrt #x# and applying the product rule gives us:

# 2{ (x)(dy/dx) + (1)(y) } +4ydy/dx = 0#

# xdy/dx + y +2ydy/dx = 0 => dy/dx = -y/(x+2y) #

Differentiating again wrt #x# and applying the product rule (twice) gives us:

# :. {(x)((d^2y)/(dx^2)) + (1)(dy/dx) } + dy/dx + 2{ (y)((d^2y)/(dx^2)) + (2dy/dx)(dy/dx)} = 0#

# :. x(d^2y)/(dx^2) + dy/dx + dy/dx + 2y(d^2y)/(dx^2) + 2(dy/dx)^2 = 0#

# :. x(d^2y)/(dx^2) + 2dy/dx + 2y(d^2y)/(dx^2) + 2(dy/dx)^2 = 0#

# :. (x+2y)(d^2y)/(dx^2) + 2(-y/(x+2y)) + 2(-y/(x+2y))^2 = 0#

# :. (x+2y)(d^2y)/(dx^2) = (2y)/(x+2y) - (2y^2)/(x+2y)^2#

# :. (d^2y)/(dx^2) = (2y)/(x+2y)^2 - (2y^2)/(x+2y)^3#

Putting over a common denominator gives:

# (d^2y)/(dx^2) = ((2y)(x+2y) - (2y^2))/(x+2y)^3#

# :. (d^2y)/(dx^2) = ((2xy+4y^2-2y^2))/(x+2y)^3#

# :. (d^2y)/(dx^2) = ((2xy+2y^2))/(x+2y)^3#

# :. (d^2y)/(dx^2) = 13/(x+2y)^3# (as # 2xy+2y^2=13 #)