Answer
Saddle points: $(0,0),(0,1),(1,0)$
Local maximum: $f( \dfrac{1}{3},\dfrac{1}{3})=\dfrac{1}{27}$
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 and local maximum or, a saddle point.
For $(x,y)=(0,0)$ $D(0,0)=-1 \lt 0$
For $(x,y)=(0,1)$ $D(0,1)=-1 \lt 0$
For $(x,y)=(1,0)$ $D(1,0)=-1 \lt 0$
For $(x,y)=( \dfrac{1}{3},\dfrac{1}{3})$
$D( \dfrac{1}{3},\dfrac{1}{3})=\dfrac{1}{3} \gt 0$ and $f_{xx} \lt 0$
Also, $f( \dfrac{1}{3},\dfrac{1}{3})=xy(1-x-y)=\dfrac{1}{27}$
Hence, Saddle points: $(0,0),(0,1),(1,0)$
Local maximum: $f( \dfrac{1}{3},\dfrac{1}{3})=\dfrac{1}{27}$