What is the extreme f(x,y)=x^2+2xy^2+(2y)/(3x) ?

The extreme f(x,y)=x^2+2xy^2+(2y)/(3x) is Saddle(-1/(2^{2/5)*3^{1/5}},-1/(\sqrt[5]{54)})

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extreme f(x,y)=x^2+2xy^2+(2y)/(3x)

extreme points

f (x, y) = x^{2} + 2 x y^{2} + \frac{2 y}{3 x}

Solution steps

Find the critical points:

\frac{\partial ^{2} f}{\partial x ^{2}} = 4 x

D (x, y) = 8 x + \frac{32 y}{3 x ^{2}} - \frac{4}{9 x ^{4}} - 16 y^{2}

Check the sign of at each critical point

Check critical point

Saddle

What is the extreme f(x,y)=x^2+2xy^2+(2y)/(3x) ?
The extreme f(x,y)=x^2+2xy^2+(2y)/(3x) is Saddle(-1/(2^{2/5)*3^{1/5}},-1/(\sqrt[5]{54)})