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$f(x,y)=x^6+2x^4-8x^2y^2+2y^4$ Has f a (strict) local minima in (0,0)?

Of course it's not a local minimum. Just take $x = y = \epsilon>0$, where $\epsilon$ is a very small number. Then $f = -4 \epsilon^4 + \epsilon^6 < f(0,0)$.
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$f(x,y)=x^6+2x^4-8x^2y^2+2y^4$ Has f a (strict) local minima in (0,0)?

You have$$f(x,x)=x^6-4x^4=x^4(x^2-4),$$which is smaller than $0$ if $x\in(-2,2)\setminus\{0\}$. But you also have $f(x,2x)=x^6+2x^4$, which is greater than $0$, unless $x=0$. Therefore, $(0,0)$ is a ...
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2 votes

$f(x,y)=x^6+2x^4-8x^2y^2+2y^4$ Has f a (strict) local minima in (0,0)?

$f(x,x)=x^6-4x^4$ which is negative in a neighborhood of $0$, but $f(\sqrt y, y)=2y^4-7y^3+2y^2$ is positive for a small $\epsilon>0$, so it's a saddle point
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1 vote

$f(x,y)=x^6+2x^4-8x^2y^2+2y^4$ Has f a (strict) local minima in (0,0)?

We know now from various answers that the origin is indeed a saddle point. The claim that the origin is not a saddle point fails because we do not need the function to alternate above and below the ...
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