Find the stationary points of the graph x^3 + y^3 = 3xy +35

Differentiate wrt x to get 3x² + 3y²dy/dx = 3y + 3x dy/dx Rearrange to get dy/dx = (3x² - 3y)/(3x-3y²). Set dy/dx =0 and infer y=x². Substitute in for y into original equation and rearrange to get x⁶ -2x³ -35 =0. Let p= x³ . Equation in terms of p becomes p² -2p -35 =0. p=7 or -5. Therefore stationary points are (7^(1/3), 7^(2/3)) and (-5^(1/3), -5^(2/3)).