Answer
$0.27$
Work Step by Step
$f_x=x \\ \implies f_x(1,1)=1$
Next, $f_y =x-6y \\ \implies f_{y}(1,1) =x-6y=1-6(1)=-5$
$f_{xx}=0; f_{yy}=-6$ and $f_{xy}=1$
We can see that the maximum of $|f_{xx}|,|f_{yy}|,|f_{xy}| $ is $6$. This implies that $M=6$
The error can be found as:
$|E(x,y,z)| \leq \dfrac{1}{2} \times (6) [ |x-1| +|y-1|)^2$
$\implies E \leq 3 \times (0.1+0.2)^2 = 0.27$
