How to study the critical points of a $2$-variable function?

Question

I am revising some past exam questions and there is one that states:

Study the critical points of the function: $$f(x,y)=x^2+y^2-x^4-y^4-2x^2y^2.$$

According to my professor, this is what I have to do:

1. Find the gradient, which is $$\nabla f(x,y)=\begin{bmatrix} 2x-4x^3-4xy^2 \\ 2y-4y^3-4x^2y \end{bmatrix}.$$

2. Find the Hessian, using partial derivatives, which is \begin{align*} f_{xx}(x,y) &= 2-12x^2-4y \\ f_{xy}(x,y) &= -8xy \\ f_{yx}(x,y) &= -8xy \\ f_{yy}(x,y) &= 2-12y^2-4x^2. \end{align*}

And until here it's all good. However, what confuses me is the next step.

We need to search at first for $Cf$, the set of critical points of $f(x,y)$, so we have to impose $\nabla f(x,y) = 0$, namely $$2x(1-2x^2-2y^2)=0 \quad \text{and} \quad 2y(1-2x^2-2y^2)=0.$$ And here I'm lost. On the answer sheet it states: $$C*f*=\{(x,y)\in\mathbb{R}^2 : x^2+y^2=1/2\} \cup \{(0,0)\}.$$ What exactly does this property above mean? Where is the $x^2+y^2=1/2$ coming from and what does $\cup \{(0,0)\}$ mean? Is $\cup \{(0,0)\}$ the critical points, which are $0$?

Please explain in the clearest and simpelst way possible!

Note: this is the first part of the full question that I am interested in. As soon as I understand this, I will proceed with answering the full question.

Thanks, guys!

Answer 1

$\cup$ means union.

From $$2x(1-2x^2-2y^2)=0 \iff 2x=0 \text{ or } 1-2x^2-2y^2=0$$ $$2y(1-2x^2-2y^2)=0 \iff 2y=0 \text{ or } 1-2x^2-2y^2=0$$

It is possible that $1-2x^2-2y^2=0 \iff x^2+y^2 = \frac12$.

Suppose not, then we must have $x=y=0$.