Find the maxima / minima of $$x^4+y^4+4x^2-xy+y^2+6$$
I tried finding the derivatives with respect to $x$ and $y$ and substituting but it did not work out since when substituting I get $8$ pairs of numbers but the solution is only the pair $(0,0)$, a global minimum.
$$\frac{\partial}{\partial x} f = 4x^3 + 8x - y$$ $$\frac{\partial}{\partial y} f = -x + 4y^3 + 2y$$ Set these both to zero to get $$4x^3 + 8x - y = 0$$ $$-x + 4y^3 + 2y = 0$$
The solution $(0,0)$ is a critical point, and if you use the extrema test, you see that it is indeed a minimum, as you found.
There is no maximum, and we know this because $f \sim O(x^4 + y^4)$, and clearly increases without bound as $x$ and $y$ increase.