Answer
$f(x,y)$ has a local minimum at $(1,1)$ and a saddle point at $(0,0)$.
Work Step by Step
Second derivative test: some noteworthy points to calculate the local minimum, local maximum and saddle point of $f$.
1. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\gt 0$ , then $f(p,q)$ is a local minimum.
2.If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \gt 0$ and $f_{xx}(p,q)\lt 0$ , then $f(p,q)$ is a local maximum.
3. If $D(p,q)=f_{xx}(p,q)f_{yy}(p,q)-[f_{xy}(p,q)]^2 \lt 0$ , then $f(p,q)$ is not a local minimum, maximum or a saddle point.
For the given question, we have $f_x(x,y)=3x^2-3y$ and $f_y(x,y)=3y^2-3x$ gives $(x,y)=(0,0)$ and $(1,1)$.
For $(x,y)=(0,0)$
$D(0,0)=-9 \lt 0$
For $(x,y)=(1,1)$
$D(1,1)=27 \gt 0$ and $f_{xx}(1,1)=6 \gt 0$
This implies that $f(x,y)$ has a local minimum at $(1,1)$ and saddle point at $(0,0)$.