How to find (implicit differetiation) dy/dx in xy=(x-y)^2?

xy = (x - y)^2
xy = x^2 - 2xy + y^2
y + xy' = 2x - (2y + 2xy') + 2yy'
y + xy' = 2x - 2y - 2xy' + 2yy'
3xy' - 2yy' = 2x - 3y
y'(3x - 2y) = 2x - 3y
y' = (2x - 3y)/(3x - 2y)

xy = (x - y)²
xy = x² - 2xy + y²
x² - 3xy + y² = 0
(d/dx)(x² - 3xy + y²) = 0
2x - 3x(dy/dx) - 3y + 2y(dy/dx)
3x(dy/dx) - 2y(dy/dx) = 2x - 3y
(3x - 2y)(dy/dx) = 2x - 3y
dy/dx = (2x - 3y)/(3x - 2y)

x y = (x-y)^2
differentiate both sides with respect to x
y + x dy/dx = 2(x-y) (1 - dy/dx)
y + x dy/dx = 2 (x- x dy/dx -y + y dy/dx)
y + x dy/dx = 2x - 2x dy/dx - 2y + 2y dy/dx

x dy/dx + 2x dy/dx - 2y dy/dx = 2x - 2y -y
dy/dx (x +2x-2y) = 2x-3y
dy/dx (3x-2y) = 2x-3y
dy/dx = (2x-3y) /(3x-2y)

xy = (x - y)^2

xy = x^2 - 2xy + y^2

x^2 - 3xy + y^2 = 0

Now differentiate using the Product Rule on the middle term:

2x - 3x(dy/dx) - 3y + 2y(dy/dx) = 0

-3x(dy/dx) + 2y(dy/dx) = 3y - 2x

(dy/dx)(-3x + 2y) = 3y - 2x

dy/dx = (3y - 2x) / (2y - 3x)