what does dy/dx look like?
6 y^2 dy +2 y dy -5 y^4 dy = 4 x^3 dx - 6 x^2 dx + 2 x dx
dy/dx = (4x^3-6x^2+2x) / (6y^2+2y-5y^4)
the curve is horizontal where dy/dx = 0
2x^3-3x^2+x = 0
so horizontal when x = 0, x = 1/2, and x = 1
when the denominator is 0, the slope is undefined (curve is vertical)
6y^2+2y-5y^4 = 0
y(6y +2 -5y^3) = 0
so jumps (hits a rock or pot hole or something) at y = 0 and possibly at three other points.
When you graph this, watch out for the points that are undefined (denominator = 0)
I don't know which "computer algebra system" you are using, but I differentiated the relation and got
y' = (4x^3 - 6x^2 + 2x)/(6y^2 + 2y - 5y^4) = 0
4x^3 - 6x^2 + 2x = 0
2x(2x^2 - 3x + 1) = 0
2x(2x-1)(x-1) = 0
x = 0 or x = 1/2 or x = 1
so you were right in part b)
(I actually volunteered to be on the evaluation group of the original MAPLE project at my old alma mater , the University of Waterloo, in Canada.
Glad to see the project worked out, was wondering what happened to it.)
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