$\text{We have,} \\ f(x,y) = x^3+3xy^2 -15x^2-15y^2+72x $

$ STEP \;1: \; \\ f_x= 3x^2+3y^2-30x + 72 \\ f_y= 6xy-30y \\ f_{xx} = 6x-30 , \; \; \; \; f_{xy}=6y , \; \; \; f_{yy}= 6x-30 $

$STEP \;2: \\ \text{We now solve, } f_x =0 , \; f_y=0,\; as \; simultaneous \; equations. \\ \therefore 3x^2+3y^2 -30 \; x +72=0 \\ i.e. \; \; x^2+y^2 -10x+24=0 \\ and \; \; 6xy-30y =0 \\ \therefore 6y(x-5)=0 $

$\\ \therefore y=0 \; \; \; or \;\;\; x=5 \\ (i) \; \; When \; y=0, \\ x^2-10x+24=0, \\ (x-6)(x-4)=0 \\ \therefore (6,0) \; (4,0) \; are \; stationary \; points. \\ (ii) When \; x=5, \\ 25+y^2-50+24=0 \\ \therefore y^2-1=0 \\ \therefore y=1 \;\;\;\;\; or \;\;\; y=\; -1 \\ \therefore (5,1) \; and \; (5,-1) \; are \; stationary \; points. $

(STEP \; 3: \ (i) \; \; When \; x=6, \; \; y=0 \ r = f_{xx} = 36-30=6 \ s= f_{xy}=0 \ t= f_{yy}=36-30=6 \ \therefore rt-s^2 = 36 \; > \; 0 \ and \; r = f_{xx}= 6 \; > \; 0 \ \therefore f(x,y) \; is \; minimum \; at \; (6,0) )

((ii) \; \; \; When \; x=4, \;\;\; y=0 \ r = f_{xx} = 24-30= \; -6 \ s= f_{xy}=0 \ t= f_{yy}=24-30=\; -6 \ \therefore rt-s^2 = 36 \; > \; 0 \ and \; r = f_{xx}= -6 \; < \; 0 \ \therefore f(x,y) \; is \; maximum \; at \; (4,0) )

$(iii) \;\;\; When \; x=5, \;\;\; y= \; 1 \\ r = f_{xx}= \; 6 \; x \; 5 -30 = 0 \\ \therefore f(x,y) \; is \; neither \; maximum \; nor \; minimum. \\ $

$(iv) \;\;\; When x=5, \;\;\; y= \; -1 \\ r = f_{xx}= \; 6 \; x \; 5 -30 = 0 \\ \therefore f(x,y) \; is \; neither \; maximum \; nor \; minimum. \\ $