Answer
Saddle point at $(0,0)$; Local maximum at $(\dfrac{-2}{3},\dfrac{2}{3})=\dfrac{170}{27}$
Work Step by Step
Given: $f_x(x,y)=3x^2-2y=0, f_y(x,y)=-3y^2-2x=0$
Simplify the given two equations.
Critical point: $(0,0)$ and $(\dfrac{-2}{3},\dfrac{2}{3})$
In order to solve this problem we will have to apply Second derivative test that suggests the following conditions to calculate the local minimum, local maximum and saddle points of $f(x,y)$ or $f(x,y,z)$.
1. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\gt 0$ , then $f(a,b)$ is a local minimum.
2. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \gt 0$ and $f_{xx}(a,b)\lt 0$ , then $f(a,b)$ is a local maximum.
3. If $D(a,b)=f_{xx}(a,b)f_{yy}(a,b)-[f_{xy}(a,b)]^2 \lt 0$ , then $f(a,b)$ is a not a local minimum or local maximum but a saddle point.
$D(0,0)=f_{xx}f_{yy}-f^2_{xy}=-4$ and $D=-4 \lt 0$
So, Saddle point at $(0,0)$
$D(\dfrac{-2}{3},\dfrac{2}{3})=f_{xx}f_{yy}-f^2_{xy}=12$ and $D=12 \gt 0$ and $f_{xx}=-4 \lt 0$
Thus, Local maximum at $(\dfrac{-2}{3},\dfrac{2}{3})=\dfrac{170}{27}$
