I like your avatar (Pythagorean theorem, eh?) so let me give one way to answer your questions.Originally Posted bymathproblem

If the tangent line is horizontal, that means its slope is zero, and y' is zero.

(x^2)*y -y^3 = 8 --------(i)

You diffferentiated it implicitly with respect to x, and you arrived at (which is correct):

y' = (-2xy)/((x^2) -3y^2)

Which is the same as,

y' = (2xy)/(3y^2 -x^2)

Set y' to zero,

0 = (2xy)/(3y^2 -x^2)

Multiply both sides by (3y^2 -x^2) /2,

0 = xy

Which means x=0, or y=0, or both x and y are zeros.

When x=0,

(x^2)*y -y^3 = 8 --------(i)

0 -y^3 = 8

y^3 =-8

y = -2

So, (0,-2) is a point of tangency for a horizontal tangent line.

When y=0,

(x^2)*y -y^3 = 8 --------(i)

0 -0 = 8

No way.

I found only one such point of tangency then.

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x^3 + y^3 - 9xy = 0 -------(ii)

You got,

dy/dx=y'=(9y -3x^2) / (3y^2 - 9x) ----which is correct also.

That will simplify to:

y' = (3y -x^2) /(y^2 -3x)

For horizontal tangent lines,

y' = 0

0 = (3y -x^2) /(y^2 -3x)

0 = 3y -x^2

x^2 = 3y

y = (x^2) /3 -----***

Substitute that into (ii),

x^3 +[(x^2) /3]^3 - 9x[(x^2) /3] = 0

x^3 +(x^6)/27 - 3x^3 = 0

(x^6)/27 -2x^3 = 0

Clear the fraction, multiplyboth sides by 27,

x^6 -54x^3 = 0

(x^3)(x^3 -54) = 0

x^3 = 0

x = 0 -------***

x^3 -54 = 0

x = cuberoot(54) = 3cbrt(2) ------***

When x=0,

x^3 + y^3 - 9xy = 0 -------(ii)

0 +y^3 -0 = 0

y = 0

So, the origin (0,0) is a point of tangency.

When x = cbrt(54),

x^3 + y^3 - 9xy = 0 -------(ii)

54 +y^3 - 9y[cbrt(54)] = 0

y^3 -(34.01786835)y +54 = 0 ------(iii)

Zeez, I will leave that to you to get the 3 roots of (iii). It's late in the evening here now.

[By "rule of signs" (?), there are 2 positive roots and one negative root.]

Therefore, there should be 4 points of tangency for horizontal tangents.

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Believe it or not, I don't know how to use any graphing calculator, any Internet "grapher", any graphing application like MathCad or what you call them.

But I can sketch any graph the old-fashion way---some points, critical points, asymptotes, known characteristics, etc.

So, since I did not sketch the graphs of your two curves, I cannot say 100% sure that the points of tangency are really points of tangency. Algebraically, or if y'=0, they are correct, though.