<s>[INST] max and min of $f(x,y)=y^8-y^4x^6+x^4$ The function is continuous in $\mathbb{R}^2$ and $f(x,y)=f(x,-y)=f(-x,y)=f(-x,-y)$.
If I consider $f(x,0)=x^4$ for $x\rightarrow +\infty$ $f$ is not limited up so $\sup f(x,y)=+\infty$.
But the origin is absolute min? 
$f(x,x)=x^8-x^{10}+x^4\rightarrow -\infty$ if $x\rightarrow \infty$? so f is not limit down
 [/INST] $f'_x=-6x^5 y^4+4x^3;\;f'_y=8y^7-4y^3x^6$
They must be both zero
$
\left\{
\begin{array}{l}
 -2 x^3 (3 x^2 y^4-2)=0 \\
 -4 y^3 \left(x^6-2 y^4\right)=0 \\
\end{array}
\right.
$
$(0,0)$ is a solution and then
$
\left\{
\begin{array}{l}
 3 x^2 y^4-2=0 \\
 x^6-2 y^4=0 \\
\end{array}
\right.
$
Solve for $y$ the first equation 
$y^4=\dfrac{2}{3x^2}$
and plug in the second
$x^6-\dfrac{4}{3x^2}=0\to x^8=\dfrac{4}{3}\to x=\pm\sqrt[8]{\dfrac{4}{3}}$
and $y^4=\dfrac{2}{3\sqrt[4]\frac{4}{3}}=\dfrac{2}{3}\sqrt[4]{\dfrac{3}{4}}=\sqrt[4]{\dfrac{4}{27}}$
$y=\pm\sqrt[16]{\dfrac{4}{27}}$
The singular points are 
$P_0(0,0);\;P_1\left(\sqrt[8]{\dfrac{4}{3}},\sqrt[16]{\dfrac{4}{27}}\right);\;P_2\left(-\sqrt[8]{\dfrac{4}{3}},\sqrt[16]{\dfrac{4}{27}}\right)$
$P_3\left(\sqrt[8]{\dfrac{4}{3}},-\sqrt[16]{\dfrac{4}{27}}\right);\;P_4\left(-\sqrt[8]{\dfrac{4}{3}},-\sqrt[16]{\dfrac{4}{27}}\right)$
To understand what is what we need second partial derivative test. Hessian matrix is:
$H(x,y)=\left(
\begin{array}{ll}
 12 x^2-30 x^4 y^4 & -24 x^5 y^3 \\
 -24 x^5 y^3 & 56 y^6-12 x^6 y^2 \\
\end{array}
\right)$
and its determinant
$D(x,y)=-24 x^2 y^2 \left(9 x^8 y^4+6 x^6+70 x^2 y^8-28 y^4\right)$
As the determinant contains all even powers of the variables and a coefficient negative, all $P_i,\;i=1,2,3,4$

Around zero $f(x,y)=x^4+O(x^5)+O(y^8)$ therefore $f$  is positive
  around the zero and the origin is a local minimum.

In the following graph we can see that $f(x,y)>0$ in a neighborhood of $(0,0)$

There is no global minimum as the function goes to $-\infty$ and no global maximum as the function goes to $+\infty$
Hope this helps
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</s>