= Solution
The <gradient> components are
$$
f_x=y(1-2x-y),\qquad f_y=x(1-x-2y).
$$
If $x=0$ or $y=0$, their simultaneous zeros give $(0,0),(0,1),(1,0)$. If $xy\ne0$, solving $2x+y=1$, $x+2y=1$ gives $(1/3,1/3)$. These are all the <critical points>. The <Hessian matrix> is
$$
H(x,y)=\begin{pmatrix}-2y&1-2x-2y\\1-2x-2y&-2x\end{pmatrix}.
$$
At each of $(0,0),(1,0),(0,1)$ its <determinant> is $-1$, so its two <eigenvalues> have opposite signs. \b[All three are saddle points of a scalar function.] At $(1/3,1/3)$,
$$
H=\begin{pmatrix}-2/3&-1/3\\-1/3&-2/3\end{pmatrix},
\qquad \operatorname{spec}(H)=\{-1,-1/3\}.
$$
\b[This point is a strict <local maximum>, of value $\boxed{1/27}$.]
The zero <level set> is exactly
$$
\boxed{\{x=0\}\ \cup\ \{y=0\}\ \cup\ \{x+y=1\}.}
$$
These three lines form a triangle containing the maximum. Inside it $f>0$; near the maximum positive <level sets> are nested closed curves, with their leading shapes elliptical because the <Hessian matrix> is negative definite. At each vertex the zero lines cross, separating alternating signs, and nearby nonzero <level sets> have hyperbolic branches. For example, the leading quadratic terms are $xy$ at $(0,0)$, $-\eta(\xi+\eta)$ for $\xi=x-1,\eta=y$ at $(1,0)$, and $-\xi(\xi+\eta)$ for $\xi=x,\eta=y-1$ at $(0,1)$.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2010/ia/paper-2-contours.png]
{title=Contours of xy(1-x-y), showing three saddle points, the interior maximum, and the three zero lines}
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