For the curve defined implicitly by x^3 + y^2−2xy=2, find the slope of the tangent line at (-1,1). Write your answer either as an integer or a reduced fraction.

Answer:Step-by-step explanation:The method you use is sort of a chain rule when you hit a y. Here's what I getd(x^3)/dx = 3x^2d(y^2)/dx = 2y * (dy/dx) This is where the chain rule comes in. -d(2xy)/dy = -2y - 2x dy/dx d(2)/dx = 0   Two is a constant. Its derivative = 0.Now add these together.3x^2 + 2y*(dy/dx) - 2y - 2x*(dy/dx) = 0Now group the terms not containing dy/dx3x^2 - 2yAnd then group those containing dy/dx2y (dy/dx) - 2x (dy/dx) Pull out the common factor 2[dy/dx]2 (dy/dx) [y - x] Put the non dy/dx terms on the right2(dy/dx) ( y + x) = - (3x^2 + 2y) Divide both sides by 2(y - x)(dy/dx) = [tex]\dfrac{-(3x^2+ 2y)}{2(-x + y)}[/tex]You have the tangent expression. dy/dx is one expression for the tangent. The other is the fraction on the right.You are almost home free. Let x = - 1Let y = 1dy/dx = -(3(-1)^2 + 2(1) ) / (2(1 + 1) dy/dx = -(3 + 2) /(2*2)dy/dx = - (5)/4dy/dx = - 1.25Check my numbers pretty carefully. The method is correct. There are so many lines that the answer may not be.