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} $

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