1. The problem statement, all variables and given/known data Find the points on the graph of x3-y3=3xy-3 where the tangent line is horizontal 2. Relevant equations y = f(x) so implicit differentiation must be used when taking the derivative of y (xy)' = xy' + y 3. The attempt at a solution So if the tangent line is horizontal the slope = 0 and the derivative of y' must be equal to zero. So first taking the derivative I end up with 3x2-3y2y' = 3xy' + 3y Factor out the three and divide by three on both sides to cancel them out. That leaves me with x2-y2y' = xy' + y Subtract y on both sides, add y2y' on both sides and get x2-y = xy' + y2y' Factor out the y' to get x2-y = y'(x+y2) divide by x+y on both sides and y' = (x2-y)/(x+y2) So at this point, I set y' equal to zero and solve for x? Don't I remove the y variables or something like that?