Using implicit differentiation to find the points on the curve?

Question

3y^3-2x^3-6x^2y+5y=0

I found dy/dx to be 6x^2+12xy/9y^2-6x^2+5, but if I set that to dy/dx to 0, I have 2 variables (x and y) to deal with. Any help would be great!

But that's what the question is asking...

Thanks Santman... very enlightening

Answer 1

9y^2*y´-6x^2-6(2xy+x^2*y´)+5y´=0
If you put y´=0
-6x^2-12xy=0 so -6x(x+2y)=0 so x= 0 which gives you
3y^3+5y=0 so y(3y^2+5) = 0 so y=0
andx=-2y
3y^3+16y^3 -24y^3+5y=0 y=0 already found and

-5y^2+5=0 y=+-1 and x= -+2

Answer 2

To find points on the curve, you should be given a pair (x,y).