Consider a special multivariate polynomial
\begin{equation*} x^4+y^8+z^2+\alpha xy^2z+\beta x^2y^4+\gamma xy^6. \end{equation*} Each term has the form $x^iy^jz^k$ satisfying $\frac{i}{4}+\frac{j}{8}+\frac{k}{2}=1$, and $(i,j,k)$ all lying on the triangle with vertices $(4,0,0),(0,8,0),(0,0,2)$.
The question is what methods can be used to determine a local extreme value at $(0,0,0)$?
My approach is to let $u=x^2,v=y^4,w=z$ which make it into a homogeneous, quadratic form \begin{equation*} u^2+v^2+w^2+\alpha u^{1/2}v^{1/2}w+\beta uv+\gamma u^{1/2}v^{3/2}, \end{equation*} but the exponents are not all integers, meaning it is not defined for negative numbers.
How to fix it, or should I use a different method? Thanks!
Compute the three partial derivatives of the original function $$F(x,y,z)=x^4+y^8+z^2+\alpha xy^2z+\beta x^2y^4+\gamma xy^6$$ $$\frac{\partial F(x,y,z)}{dx}=4 x^3+2 \beta x y^4+\gamma y^6+\alpha y^2 z\tag1$$ $$\frac{\partial F(x,y,z)}{dy}=4 \beta x^2 y^3+6 \gamma x y^5+2 \alpha x y z+8 y^7\tag2$$ $$\frac{\partial F(x,y,z)}{dz}=\alpha x y^2+2 z\tag3$$
Use $(3)$ to get $$z=-\frac{1}{2} \alpha x y^2$$ Plug it in $(1)$
$$4 x^3-\frac{y^4 \left(\alpha ^2-4 \beta \right)}{2} x +\gamma y^6=0$$ which is a cubic equation in $x$.
Once solved, replace $x$ in $(2)$ and solve for $y$ $$y^3\left( 8 y^4+6 \gamma x y^2-x^2 \left(\alpha ^2-4 \beta \right)\right)=0$$ where $x$ is a function of $y$.
Have fun for the last part.