differentiation

(a) x^3+y^3-3xy+1=0
Differentiate w.r.t. x
3x^2 + 3y^2 * dy/dx - 3x*dy/dx - 3y + 0 = 0
dy/dx (3y^2 - 3x) = 3y - 3x^2
dy/dx = (y-x^2) / (y^2 - x)
(b) Let the required point on C be (a, b)
Slope of tangent at that point = dy/dx at (a,b) = (b-a^2) /(b^2 - a) = 0
(as slope of x-axis = 0)
b-a^2 = 0
b = a^2
Also, (a, b) lies on C, i.e. a^3 + b^3 -3ab + 1 = 0
Hence a^3 + (^2)^3 - 3a(a^2) + 1 = 0
a^6 - 2a^3 + 1 = 0
(a^3 - 1) = 0
a^3 = 1
a= 1
b = 1^2 = 1
The point is (1, 1)

a) 3x^2 + 3y^2 (dy/dx) - 3y - 3x (dy/dx) = 0
搬黎搬去,, 變成...(y-x^2)/(y^2-x)

b) if the point on C at which the tangent to C is parallel to the x-axis,
dy/dx = 0
(y-x^2)/(y^2-x) = 0
y = x^2 or y^2 = x
y = 1, x = 1 or y=0, x=0 (rej)