## Calculus (3rd Edition)

$(2^{\frac{1}{3}}, 2^{\frac{2}{3}})$
The tangent line is horizontal when $\frac{dy}{dx}=0$ Implicit differentiation of $x^{3}+y^{3}=3xy$ gives $3x^{2}+3y^{2}\frac{dy}{dx}=3(y+x\frac{dy}{dx})$ or $x^{2}-y=x\frac{dy}{dx}-y^{2}\frac{dy}{dx}$ Or $\frac{dy}{dx}=\frac{x^{2}-y}{x-y^{2}}$ $\frac{dy}{dx}=0\implies x^{2}-y=0\implies y=x^{2}$ Substituting $y=x^{2}$ in $x^{3}+y^{3}=3xy$, we get $x^{3}+(x^{2})^{3}=3x(x^{2}$). That is, $x^{3}+x^{6}=3x^{3}$ $x=0$ satisfies the equation and in that case $y=0^{2}=0$. The point (0,0), which is the origin, is the one point where the tangent line is horizontal. But, we need to find the other point. $x^{3}+x^{6}=3x^{3}\implies x^{6}=2x^{3}$ $\implies x^{3}=2$ or $x=2^{1/3}$ $y=x^{2}=(2^{\frac{1}{3}})^{2}=2^{2/3}$ So, the other point is $(2^{\frac{1}{3}}, 2^{\frac{2}{3}})$.