We have$ f(x, y) = x^{3} y^{3}(1-x-y)$

STEP 1 :$ f_{x} = y^{2}[3x^{2}(1-x-y)-x^{3}]$

=  $y^{2}(3x^{2} – 4x^{3} – 3x^{2}y)$

=$ 3x^{2}y^{2} – 4x^{3}y^{2} – 3x^{2}y^{3}$

$f_{x} = x^ {3}[2y(1-x-y)-y^{2}]$

= $x^{3}(2y-2xy-3y^{2})$

= $2x^{3}y – 2x^{4}y – 3x^{3}y^{2}$

$f_{xx} = 6xy^{2} – 12x^{2}y^{2} – 6xy^{3},$

$f_{xy} = 6x^{2}y - 8x^{3}y - 9x^{2}y^{2},$$f_{yy} = 2x^{3} - 2x^{4} - 6x^{3}y

$ STEP 2: We solve $ f_x = 0, f_y = 0$ simultaneously. $\therefore 3x^{2}y^{2} – 4x^{3}y^{2} – 3x^{2}y^{3} = 0$ $x^{2}y^{2}(3-4x-3y) = 0$ and $2x^{3}y – 2x^{4}y – 3x^{3}y^{2} = 0 $                                                        $x^{3}y(2-2x-3y) = 0$ $\therefore x=0, y=0$ and $3-4x-3y = 0, 2-2x-3y = 0$ Subtracting, we get 1 – 2x = 0 $\therefore x = \dfrac{1}{2}$ $\therefore = 3y = 3-4(\dfrac{1}{2}) = 1$ $\therefore y = \dfrac{1}{3}$ Hence, the stationary points are (0,0) and $(\dfrac{1}{2},\dfrac{1}{3})$  STEP 3: 1.      At x = 0, y = 0, r = 0, s = 0, t = 0 $\therefore rt – s^{2} = 0$ Hence, our method fails and we reject this pair. At x = $(\dfrac{1}{2})$, y =$( \dfrac{1}{3}) $                                          r =$ f_{xx} = 6(\dfrac{1}{2},\dfrac{1}{9}) - 12(\dfrac{1}{4},\dfrac{1}{9}) - 6(\dfrac{1}{2},\dfrac{1}{27})$ =$ (\dfrac{1}{3} - \dfrac{1}{3}) - (\dfrac{1}{9}) = -(\dfrac{1}{9})$   s =  $f_{xy} = 6(\dfrac{1}{4},\dfrac{1}{3}) - 8(\dfrac{1}{8},\dfrac{1}{3}) - 9(\dfrac{1}{4},\dfrac{1}{9})$ = $(\dfrac{1}{2} - \dfrac{1}{3} - \dfrac{1}{4}) = -(\dfrac{1}{12})$   t = $ f_{yy} = 2(\dfrac{1}{8}) - 2(\dfrac{1}{16}) - 6(\dfrac{1}{8},\dfrac{1}{3})$ = $(\dfrac{1}{4} - \dfrac{1}{8}- \dfrac{1}{4}) = -(\dfrac{1}{8})$   rt – $s^{2} = (\dfrac{-1}{9})( \dfrac{-1}{8}) - (\dfrac{1}{12})^{2}$ =\((\dfrac{1}{72} - \dfrac{1}{144}) = (\dfrac{1}{144}) \gt 0\) And r =\( (\dfrac{-1}{9}) \lt 0\) \therefore f(x, y) is a maxima.   Maximum value = $\dfrac{1}{8}\cdot\dfrac{1}{9} (1-\dfrac{1}{2} - \dfrac{1}{3})$ =$ \dfrac{1}{72}\cdot\dfrac{1}{6} = \dfrac{1}{432}$