Question 401556
{{{f(x,y) = xy(1-x-y)   = xy - x^2y-xy^2}}}

==> {{{f[x] = y-2xy - y^2 = y(1-2x-y) = 0}}}, and 
{{{f[y] = x-2xy - x^2 = x(1-x-2y) = 0}}}.
Solving the preceding system above, we get the following c.p.'s
(i) (0,0)
{ii) (1,0)
(iii) (0,1)
(iv) (1/3, 1/3)
Finding the 2nd order partial derivatives:
{{{f[xx] = -2y}}}
{{{f[yy] = -2x}}}, and 
{{{f[xy] = 1-2x - 2y = f[yx]}}}
(i) At (0,0), {{{f[xx] = 0}}}
{{{f[yy] = 0}}}, and {{{f[xy] = 1}}}
D = {{{f[xx]* f[yy] - (f[xy])^2 = 0 * 0 - 1 = -1 < 0}}}.
Therefore there is a saddle point at (0,0).


(ii) At (1,0), {{{f[xx] = 0}}}
{{{f[yy] = -2}}}, and {{{f[xy] = -1}}}
D = {{{f[xx]* f[yy] - (f[xy])^2 = 0 * -2  - 1 = -1 < 0}}}.
Therefore there is a saddle point at (1,0)


(iii) At (0,1), {{{f[xx] = -2}}}
{{{f[yy] = 0}}}, and {{{f[xy] = -1}}}
D = {{{f[xx]* f[yy] - (f[xy])^2 = -2 * 0 - 1 = -1 < 0}}}.
Therefore there is saddle point at (0,1).


(iv) At (1/3, 1/3), {{{f[xx]= -2/3}}}
{{{f[yy] = -2/3}}}, and {{{f[xy] = -1/3}}}
D = {{{f[xx]* f[yy] - (f[xy])^2 = (-2/3)^2 - 1/9 = 4/9-1/9 = 1/3 > 0}}}.
Therefore there is a relative maximum at (1/3, 1/3).