Answer
$$ 0.27$$
Work Step by Step
$f_x(1,1) =x =1$
and $f_y(1,1) =x-6y =1-(6)(1)=-5$
Now $$f_{xx}=0 \\ f_{yy}=-6 \\ f_{xy}=1$$
We notice that the maximum of $|f_{xx}|,|f_{yy}|,|f_{xy}| $ is $6$. So, $M=6$
Error:
$|E(x,y,z)| \leq (\dfrac{1}{2}) (6) [ |x-1| +|y-1|)^2$
or, $$ E \leq (3) (0.1+0.2)^2 = 0.27$$
