Simple question - How $x^4 +y^4 -2(x-y)^2$ having a minimum at $(0,0)$ I know this is a silly question, but I need to prove that
$$ f(x,y) = x^4 + y^4 -2 (x-y)^2 $$
is having a minima at $(0,0)$.
I tried doing maxima and minima method to it but the determinant i.e $AC- B^2= 0 $ 
So, we need to prove it algebraically only.

$f_x = 4x^3 - 4x +4y = 0, f_y = 4y^3 - 4y + 4x = 0\implies 4(x^3+y^3) = 0 \implies x = -y$ or $x^2 - xy + y^2 = 0$ which yields $x = y = 0$. If $x = -y \implies 4x^3 - 8x = 0 \implies 4x(x^2 - 2) = 0\implies x = 0, \pm \sqrt{2}\implies y = 0,\mp \sqrt{2}$. Next $f_{xx}= 12x^2 - 4$. Thus $f_{xx}(\pm \sqrt{2}) = 20 > 0$. So by the second derivative test ( see a calculus book for this test ) the min are at these points. Thus $f_{\text{min}} = f(\pm \sqrt{2}, \mp \sqrt{2})= -8 $ since $f(0,0) = 0 > -8$.
