SOLUTION: Hi there! Help would be super appreciated. This is for Calculus homework, but idk which category this would fit under. Find all points on the graph of y^3-27y=x^2-90 at which

Differentiate both sides.  You will get


    3y^2*y' - 27y' = 2x


From this equation express y', the derivative of y :


    y' = .


From this formula, y' is vertical there, where the denominator is zero:

        3y^2 - 27 = 0,


which implies  

       y^2 = 27/3 = 9;  

       y = +/- 3.


Substitute these values of y into your original equation to find x.  You will get two cases:


    a)  for y = 3    3^3 - 27*3 = x^2 - 90

                     27 - 27*3  = x^2 - 90

                     -54        = x^2 - 90

                     -54 + 90 = x^2

                     x^2 = 36

                     x = +/- 6.


        So, you get two pairs of points  (x,y) = (6,3)  and (-6,3)


Similarly analyze the other case  y = -3.