Solve the equation $F(x,y) =2x^3 - 2y^3 + 3x^2y + 3y$ and define the stationary points

Question

I am kind of lost as how to solve this question. I have to derive it first to get the stationary points by putting the derivative equation $=0 $, but here I have two variables, what do I do ?

$F (x,y) =2x^3 - 2y^3 + 3x^2y + 3y$?

Answer 1

I got a maximum for $$(x,y)=(-1,1)$$ saddle points for $$(0,\pm \frac{1}{\sqrt{2}})$$ and a minimum for $$(1;-1)$$

Answer 2

• $\frac{\partial F}{\partial x}=6x^2+6xy=6x(x+y)=0\Rightarrow x=0$ or $x=-y$
• $\frac{\partial F}{\partial y}=-6y^2+3x^2+3=0$. If $x=0$ then $y=\pm\frac{\sqrt{2}}{2}$, so $(0,\pm\frac{\sqrt{2}}{2})$. If $x=-y$ then $(\pm 1,\mp 1)$.