Find the critical points of the following functions. Use the Second Derivative Test to determine whether each critical point corresponds to a loal maximum, local minimum, or saddle point. f(x,y)=x^4+2y^2-4xy

Let ${\displaystyle c=-(a,b)}$ point of the given function. Then we have
${\displaystyle \frac{\partial (x^4+2x^2-4xy)}{\partial x}{\mid }_{a,b}=0}$ and ${\displaystyle \frac{\partial (x^4+2x^2-4xy)}{\partial y}{\mid }_{a,b}=0}$
which further implies that ${\displaystyle (4x^3-4y){\mid }_{a,b}=0}$ and ${\displaystyle (4y-4x){\mid }_{a,b}=0}$ implying that ${\displaystyle a^3-b=0}$ and b-a=0 leading to ${\displaystyle (a,b)\equiv (0,0),(1,1),(-1,-1).}$
Therefore the required critical points for the given function are (0, 0), (1, 1), and (-1, -1).
Now we calculate the values of ${\displaystyle \frac{{\partial }^{2}f}{\partial x^2},\frac{{\partial }^{2}f}{\partial y^2}}$ and ${\displaystyle \frac{{\partial }^{2}f}{\partial x\partial y}}$ to be able to use the second derivative test. We get ${\displaystyle \frac{{\partial }^{2}f}{\partial x^2}=\frac{\partial }{\partial x}(4x^3-4y),\frac{{\partial }^{2}f}{\partial y^2}=\frac{\partial }{\partial y}(4y-4x)}$, and ${\displaystyle \frac{{\partial }^{2}f}{\partial x\partial y}=\frac{\partial }{\partial x}}$ mplying that ${\displaystyle \frac{{\partial }^{2}f}{\partial x^2}=12x^2,\frac{{\partial }^{2}f}{\partial y^2}=4}$, and ${\displaystyle \frac{{\partial }^{2}f}{\partial x\partial y}=-4}$.
Now, we denote ${\displaystyle D\left(c\right)=\left(12x^2{\mid }_c\right)\left(4{\mid }_c\right)-{\left((-4){\mid }_c\right)}^2}$
Now we calculate this for all the above three critical points we found as
${\displaystyle D(0,0)=(12{\left(0\right)}^{24}-{(-4)}^2=-16).}$
${\displaystyle D(1,1)=(12{\left(1\right)}^{24}-{(-4)}^2=31).}$
${\displaystyle D(-1,-1)=(12{(-1)}^{24}-{(-4)}^2=31).}$
We also have

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