We have $f(x,y) = x^3 + y^3 - 3axy$

Step1:

$f_x = 3x^2 - 3ay, \ f_y = 3y^2 - 3ax$

$r=f_{xx} = 6x, \ t=f_{yy} = 6y, \ s=f_{xy} = -3a$

Step 2:

We now solve, $f_x = 0, \ f_y =0$ as a simultaneous equations.

$\therefore x^2 - ay = 0 $ and $y^2 - ax = 0$

To eliminate y, we put y=x2/a in second equation

$\therefore x^4 - a^3 x=0$

$\therefore x(x^3 - a^3) = 0$

$\therefore x = 0 \text { or } x =a$

When x=0, y=0 and when x=a, y=a

∴ (0,0) and (a,a) are stationary points.

Step 3:

(i) For $x=0, \ y=0, \ r=f_{xx} =0, \ t=f_{yy} = 0, \ s = f_{xy} = -3a$

Hence, (r \ t-s^2 = 0 -9a^2 < 0)

We reject this pair.

(ii) For $x=a, \ y=a, \ r=f_{xx} = 6a, \ t=f_{yy} = 6a, \ s=f_{xy} = -3a$

Hence, (r\ t - s^2 = 36a^2 - 9a^2 = 27 a^2 > 0)

∴ f(x,y) is stationary at x=a, y=a.

But, (r= f_{xx} = 6a>0, \ \text{ since } a>0 )

∴ f(x,y) is stationary at x=a, y=a.

Putting $x=a, \ y =a $in $x^3 + y^3 -3axy$

The minimum value of

$\therefore f(x,y) = a^3 + a^3 -3a^3$

$\therefore f(x, y) = -a^3$