f(x,y)=x²-xy+y²/2+3
The question seems incomplete, however, I can guess that you are asking to find the linearization of "f(x,y)=x^2-xy+y^2\/2+3" at a given point say (3,2). If this is the case
"f(3,2)=9-6+2+3=8"
"\\frac{\\partial \\:}{\\partial \\:x}\\left(f\\left(x,y\\right)\\right)=2x-y"
"\\frac{\\partial \\:}{\\partial \\:y}\\left(f\\left(x,y\\right)\\right)=-x+y"
"\\nabla f(x,y)=<2x-y,-x+y>"
"\\nabla f(x,y)=<4,-1>"
Linearization "z-8=4(x-3)-1(y-2)"
"4x-y-12+2-z+8=0"
"4x-y-z-2=0"
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