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