Differentiate implicitly:
\(3x^2-y-xy'+3y^2y'=0\)
Solve for y
\(y'(3y^2-x)=y-3x^2
\\
\\y'={y-3x^2\over3y^2-x}\)
When the tangent is parallel to the x-axis we have y'=0, so we must solve
\(y'={y-3x^2\over3y^2-x}=0\implies y=3x^2\)
To find the actual value of x we plug this expression for y into the original equation
\(x^3-3x^3+27x^6=0
\\
\\x^3(27x^3-2)=0\implies x=\{0,{\sqrt[3]2\over3}\}\)
Plugging this into the formula for y above gives the points
\((0,0)\text{ and }({\sqrt[3]2\over3},{\sqrt[3]4\over3})\)
which is where our tangent will be parallel to the x-axis.