Find critical points of the function $x^2+4xy+4y^2+x^3+2x^2y+y^4$ 
Find critical points of the function $x^2+4xy+4y^2+x^3+2x^2y+y^4$

I found that $(0,0)$ is a critical point but I couldn't able to define it whether it'll be maximum or minimum. What are the other critical points of the function? What should I do now?
 A: Consider a function $f(x,y)$. For a point $(a,b)$ to be a critical point,the  partial derivatives at that point have to zero. 
$$ \partial_x f(x,y) = 0 ,\partial_y f(x,y) = 0$$
Where $\partial_{x_i} f(x_1, x_2, ..) = \cfrac{\partial f(x_1, x_2..)}{\partial x_i} $ 
Define $$ D = 
        \begin{vmatrix}
     \partial_x^2 f(x,y)& \partial_{xy} f(x,y) \\
     \partial_{yx} f(x,y) & \partial_y^2 f(x,y) 
         \end{vmatrix} 
        $$
Now, If a point, say $P(a,b)$ is a critical point then there are three possible cases:


*

*Local Maximum: If $D \gt 0$ and $\partial_x^2 f(x,y) < 0$

*Local Minimum: If $D \gt 0$ and $\partial_x^2 f(x,y) > 0$

*Saddle point:  If $D < 0$
Now, having knowledge of these, I hope you can solve your problem,now.
A: Regarding the nature of the critical point $(0,0)$:
The Hessian matrix at that point represents the quadratic part of the power series for the expression at that point, which is $x^2+4xy+4y^2=(x+2y)^2$. That would make $(0,0)$ a degenerate critical point because that expression has its minimum value $0$ at every point of the line $x+2y=0$.
Also including the cubic part of the original expression gives $x^2+4xy+4y^2+x^3+2x^2y=(x+2y)(x+x^2+2y)$, which for any fixed value of $x$ is negative when $-x^2-x<2y<-x$ with a minimum value of $-\frac{x^4}4$ where $2y=-x-\frac{x^2}2$. That would make $(0,0)$ a saddle point.
So the final question is wether also including the quartic part $y^4$ will still leave points with negative values arbitrarily close to $(0,0)$ or wether it will turn $(0,0)$ into a local minimum.
