Step 1: Differentiate the given curve

The given equation is: $x^3 + y^3 = 6xy$

To find the slope of the tangent line, differentiate both sides with respect to $x$, remembering to apply the chain rule to terms involving $y$: $\frac{d}{dx}(x^3) + \frac{d}{dx}(y^3) = \frac{d}{dx}(6xy)$

This simplifies to: $3x^2 + 3y^2 \frac{dy}{dx} = 6y + 6x \frac{dy}{dx}$

Rearrange terms to isolate $\frac{dy}{dx}$: $3y^2 \frac{dy}{dx} - 6x \frac{dy}{dx} = 6y - 3x^2$

Factor out $\frac{dy}{dx}$: $\frac{dy}{dx} (3y^2 - 6x) = 6y - 3x^2$

Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}$

Step 2: Slope at the point (3, 3)

Substitute $x = 3$ and $y = 3$ into the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{6(3) - 3(3^2)}{3(3^2) - 6(3)}$

Simplify the numerator and denominator: $\frac{dy}{dx} = \frac{18 - 27}{27 - 18} = \frac{-9}{9} = -1$

The slope of the tangent line at $(3, 3)$ is $-1$.

Step 3: Equation of the tangent line

The equation of a line is: $y - y_1 = m(x - x_1)$

Here, $m = -1$ and $(x_1, y_1) = (3, 3)$: $y - 3 = -1(x - 3)$

Simplify: $y - 3 = -x + 3$

$y = -x + 6$

Thus, the equation of the tangent line at $(3, 3)$ is: $y = -x + 6$

Step 4: Tangent line is horizontal

The tangent line is horizontal when $\frac{dy}{dx} = 0$. From the expression for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x}$

Set the numerator equal to 0: $6y - 3x^2 = 0$

Solve for $y$: $y = \frac{x^2}{2}$

Now substitute $y = \frac{x^2}{2}$ into the original equation to find $x$: $x^3 + \left(\frac{x^2}{2}\right)^3 = 6x\left(\frac{x^2}{2}\right)$

Simplify: $x^3 + \frac{x^6}{8} = 3x^3$

Multiply through by 8 to eliminate the fraction: $8x^3 + x^6 = 24x^3$

Rearrange: $x^6 - 16x^3 = 0$

Factorize: $x^3(x^3 - 16) = 0$

Thus: $x^3 = 0 \quad \text{or} \quad x^3 = 16$

Since $x > 0$ in the first quadrant, $x^3 = 16$, so: $x = \sqrt[3]{16}$

Substitute $x = \sqrt[3]{16}$ into $y = \frac{x^2}{2}$: $y = \frac{(\sqrt[3]{16})^2}{2}$

Thus, the point where the tangent line is horizontal in the first quadrant is: [ \left(\sqrt[3]{16}, \frac{(\sqrt[3]{16})^2}{2}\right) ]

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Final Answers:

1. Equation of the tangent line at $(3, 3)$: $y = -x + 6$

2. Point where the tangent line is horizontal in the first quadrant: [ \left(\sqrt[3]{16}, \frac{(\sqrt[3]{16})^2}{2}\right) ]

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Related Questions:

1. How is implicit differentiation used to solve problems involving curves?
2. Can tangent lines have undefined slopes on certain curves?
3. How does the second derivative determine concavity at a given point?
4. What is the geometric interpretation of $\frac{dy}{dx}$?
5. How can tangent lines be applied in optimization problems?

Tip:

Always check if a curve is symmetric, as it can simplify finding critical points and tangent line behaviors.