Find the points on the graph of x3-y3=3xy-3 where the tangent line is horizontal
y = f(x) so implicit differentiation must be used when taking the derivative of y
(xy)' = xy' + y
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?