Answer
$(0,0)$ is a saddle point and $(-1,-1)$ is a local maximum.
Work Step by Step
Let us first find the critical points of the function where $f_x$ and $f_y$ are equal to 0, assuming first and second partial derivatives are continuous.
$f_x=3x^2+3y$ and $f_y=3y^2+3x$
Let us substitute $x=-y^2$ from the second equation into the first equation. $f_x=3y^4+3y=0$
$f_x=3y^4+3y=3y(y^3+1)=0$
$y=0$ and $y=-1$
Therefore, we have the following two critical points, $(0,0)$ and $(-1,-1)$.
$f_{xx}=6x$ and $f_{yy}=6y$ and $f_{xy}=3$
$D(0,0)=(0)(0)-(3)^2<0$, which means $(0,0)$ is a saddle point.
$D(-1,-1)=(-6)(-6)-3^2>0$ and $f_{xx}<0$, which means the point $(-1,-1)$ is a local maximum.
$f(-1,-1)=1$
