Originally Posted by
Chris L T521 The function is $\displaystyle f\!\left(x,y\right)=x^4-8xy+2y^2-3$
Thus, $\displaystyle \frac{\partial f}{\partial x}=4x^3-8y$ and $\displaystyle \frac{\partial f}{\partial y}=4y-8x$
Now to find the critical points, solve $\displaystyle \frac{\partial f}{\partial x}=0$ and $\displaystyle \frac{\partial f}{\partial y}=0$
This would mean that $\displaystyle 4x^3-8y=0$ and $\displaystyle 4y-8x=0$.
Solving the second equation, we have $\displaystyle 4y=8x$
Substituting this into the first equation, we have $\displaystyle 4x^3-2(8x)=0\implies 4x^3-16x=0\implies 4x(x^2-4)=0$
Thus, $\displaystyle x=0,~x=2$ or $\displaystyle x=-2$.
As a result, when:
$\displaystyle x=0\,:\, 4y=8(0)\implies y=0$
$\displaystyle x=2\,:\, 4y=8(2)\implies y=4$
$\displaystyle x=-2\,:\, 4y=8(-2)\implies y=-4$
Thus, your critical points should be $\displaystyle \left(0,0\right)$, $\displaystyle \left(2,4\right)$ and $\displaystyle \left(-2,-4\right)$
Does this make sense?