Question

f(x,y)=x²-xy+y²/2+3

Answer 1

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 $z-8=4(x-3)-1(y-2)$

$4x-y-12+2-z+8=0$

$4x-y-z-2=0$