A curve C has the following equation: x^3 + 3y - 4(x^3)*(y^3) a) Show that (1,1) lies on C b) Find dy/dx

a) Substituting the coordinate (1,1) into the left hand side of the equation for C we obtain: (1³) + 3*1 - 4(1³)(1³) = 1 + 3 - 4 = 0 = The right hand side of the equation, hence the equation is satisfied, and therefore (1,1) lies on C.

b) Differentiating implicity we find: 
3x² + 3dy/dx - 12x²y³ - 12x³y²dy/dx = 0
Rearranging yields:
3x² - 12x²y³= (12x³y² - 3)dy/dx

Hence dy/dx = (3x² - 12x²y³/(12x³y² - 3)
Which simplifies to 

dy/dx = x²(1 - 4y³)/(4x³y² - 1)

(An alternative expression can be obtained be moving the terms not involving dy/dx to the right hand side)