Equation given is
"f(x,y) = 6x^{2} - 9x-3xy-7y+5y^{2}"
taking partial derivative with respect to x and y
"\\frac{\\partial f}{\\partial x} = 12x-9-3y"
"\\frac{\\partial f}{\\partial y} = -3x-7+10y"
solving above equation to find x and y
"12x-9-3y=0" "\\implies y = 4x-3" . . . . . . . (i)
"-3x-7+10y=0" . . . . . . . (ii)
putting value of y from (i) to (ii),
"-3x-7+10(4x-3) = 0"
"-3x-7 + 40x -30 = 0 \\implies x=1"
then from (i),
"y=4(1) - 3 = 1"
taking higher partial derivatives,
"r =\\frac{\\partial^{2} f}{\\partial x^{2} } = 12"
"t= \\frac{\\partial^{2} f}{\\partial y^{2} } = 10"
"s= \\frac{\\partial^{2} f}{\\partial x \\partial y } = -3"
Now
"rt-s^{2} = 12*10 - 9 > 0"
So point (1,1) is point of local minima.
Minimum value of f(x,y)
"f(1,1) = 6-9-3-7+5 = -8"
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