Answer
(a) $\frac{5}{4}$, $\frac{4}{5}$
(b) $(3\sqrt[3] 2, 3\sqrt[3] 4)$
(c) $A(3\sqrt[3] 4, 3\sqrt[3] 2)$
Work Step by Step
(a) Given the equation $x^3+y^3-9xy=0$, differentiate both sides with respect to $3x^2+3y^2y'-9y-9xy'=0$ which gives $y'=\frac{x^2-3y}{3x-y^2}$
Thus the slope of the tangent line at the given point $(4,2)$ is $m_1=y'=\frac{4^2-3(2)}{3(4)-2^2}=\frac{5}{4}$ and the slope of the tangent line at the given point $(2,4)$ is $m_2=y'=\frac{2^2-3(4)}{3(2)-4^2}=\frac{4}{5}$
(b) For a horizontal tangent, $y'=0$, $\frac{x^2-3y}{3x-y^2}=0$, we have $y=x^2/3$. Use it in the original equation, $x^3+(x^2/3)^3-9x(x^2/3)=0$; discard solution $x=0$ and we have $x^3/27=2$, which gives $x=3\sqrt[3] 2$ and $y=x^2/3=3\sqrt[3] 4$. That is, point $(3\sqrt[3] 2, 3\sqrt[3] 4)$
(c) For the vertical tangent at point A, the slope will be $\infty$ which requires that $3x-y^2=0$ or $x=y^2/3$.
Use it in the original equation: $(y^2/3)^3+y^3-9(y^3/3)y=0$. Discard the solution $y=0$. Thus, we have $y^3=54$ and $y=3\sqrt[3] 2$ and $x=y^2/3=3\sqrt[3] 4$. That is, $A(3\sqrt[3] 4, 3\sqrt[3] 2)$
