Answer
$(2^{\frac{1}{3}}, 2^{\frac{2}{3}})$
Work Step by Step
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 $(2^{\frac{1}{3}}, 2^{\frac{2}{3}})$.
