Folium of Descartes

Find the local extreme points for local functions defined by implicit function theorem for
$x^3+y^3-3xy=0$
We have $3x^2+3y^2(x)y'(x)-3y(x)-3xy'(x)=0$
From this relation we have

$y'(x)=\frac{y(x)-x^2}{y^2(x)-x}$

Extreme points are between solutions of $x^3+y^3(x)-3xy(x)=0$ and y'(x)=0 ,or $y(x)-x^2=0$
so $x^3+x^6-3x^3=0\; x^6-2x^3=x^3(x^3-2)=0$
Solution x=0 is of no interest as near point (0,0) implicit function theorem is not applicable.
We have $x=\sqrt[3]{2}$ and $y(x)=y(\sqrt[3]{2})=\sqrt[3]{4}$
Now

$y"(x)=\frac{(y'(x)-2x)(y^2(x)-x)-(y(x)-x^2)(2y(x)y'(x)-1)}{(y^2(x)-x)^2}$

For $x=\sqrt[3]{2}$ we have $y(x)-x^2=0$ so

$y"(\sqrt[3]{2})=\frac{-2\sqrt[3]{2} }{\sqrt[3]{8}-\sqrt[3]{2} }=-2$

As second derivative is negative in $x=\sqrt[3]{2}$ implicit function has a local maximum point in $x=\sqrt[3]{2}$ with value $y(x)=y(\sqrt[3]{2})=\sqrt[3]{4}$