You need to use implicit differentiation, such that:

`(d(xy))/(dx) = (d(1–x–y)^2)/(dx)`

Using the product rule to the left and chain rule to the right, yields:

`(dx)/(dx)*y + x*(dy)/(dx) = 2(1 - x - y)*(d(1 - x - y))/(dx)`

`y + x*(dy)/(dx) = 2(1 - x - y)*(0 - 1 - (dy)/(dx))`

`y + x*(dy)/(dx) = -(2 - 2x - 2y)(1 + (dy)/(dx))`

`y + x*(dy)/(dx) = (2x + 2y - 2)(1 + (dy)/(dx) )`

`y + x*(dy)/(dx) = 2x + 2x(dy)/(dx) + 2y + 2y(dy)/(dx) - 2 - 2(dy)/(dx)`

You need to isolate to one side the terms that contain (dy)/(dx) such that:

`x*(dy)/(dx) - 2x(dy)/(dx) - 2y(dy)/(dx) + 2(dy)/(dx) = 2x + y - 2`

Factoring out `(dy)/(dx)` yields:

`(dy)/(dx)(2 - x - 2y) = 2x + y - 2 => (dy)/(dx) = (2x + y - 2)/(2 - x - 2y)`

**Hence, evaluating the derivative of the given function, using implicit differentiation, yields **`(dy)/(dx) = (2x + y - 2)/(2 - x - 2y).`

xy = (1-x-y)^2.

To find the derivative of xy = (1-x-y)^2.

Differentiating both sodes with respect to x we get:

(xy)' = {(1-x-y)^2}'.

xy' +x'y = 2(1-x-y)^(2-1) * (1-x-y)'

xy'+1*y = 2(1-x-y)(0-1-y')

xy' +y = -2(1-x-y) - 2(1-x-y)y'

We solve for y':

xy'+2(1-x-y)y' = -2(1-x-y) -y = -2+2x+2y-y

y' (x+2-2x-2y) = 2x+y-2

y'(2-x-2y) = (2x+y-2)

y' = (2x+y-2)/(2-x-2y) .