Let $f : \Bbb R^2 \longrightarrow \Bbb R$ be defined by $$f(x,y) = x^6 - 2x^2y -x^4y +2y^2.$$
( $\Bbb R$ is the set of all real numbers and $\Bbb R^2 = \left \{(x,y)\ :\ x,y \in \Bbb R \right \}$ )
Which one of the following statements is TRUE?
$(\text A)$ $f$ has a local maximum at the origin.
$(\text B)$ $f$ has a local minimum at the origin.
$(\text C)$ $f$ has a saddle point at the origin.
$(\text D)$ The origin is not a critical point of $f.$
I find that $\frac {\partial f} {\partial x}\ \bigg |_{(0,0)} = \frac {\partial f} {\partial y}\ \bigg |_{(0,0)} = 0$ and $\Delta (0,0) = \frac {{\partial}^2 f} {\partial x \partial y}\ \bigg |_{(0,0)} - \frac {{\partial}^2 f} {\partial x^2}\ \bigg |_{(0,0)} \cdot \frac {{\partial}^2 f} {\partial y^2}\ \bigg |_{(0,0)} = 0.$ So $(0,0)$ is definitely a critical point of $f$ but no further conclusion can further be made as $\Delta (0,0) = 0.$ So how to determine the nature of the point $(0,0)$ for the function $f$? Any suggestion regarding this will be highly appreciated.
Thank you very much for your valuable time.
Hint. Note that $$f(x,y) = x^6 - 2x^2y -x^4y +2y^2=(y-x^2)\cdot(2y-x^4).$$ Does the sign of $f$ change in any neighbourhood of the origin (where $f$ attains the value $0$)?
Consider for example the behaviour of the sign of $f$ along the curve $y=x^3$: $$f(x,x^3) = (x^3-x^2)\cdot(2x^3-x^4)=x^5(x-1)\cdot(2-x).$$ Can you take it from here?