I will use the following example to show you how to do this quesiton.
Question
Use the linear approximation to estimate $(0.98)^2(-1.01)^3$, and compare with the value given by a calculator and compute the percentage error.
Solution
Let $f(x,y)=x^2y^3$. The linear approximation of $f(x,y)$ at $(1,-1)$ (we take the integral coordinate that is closed to $(0.98,-1.01)$) is
\[L(x,y)=f(1,-1)+\frac{\partial f}{\partial x}(1,-1)(x-1)+\frac{\partial f}{\partial y}(y-(-1)).\]After computation, we have
\[L(0.98,-1.01)=-1+2\cdot(1)\cdot(-1)^3\cdot(0.98-1) + 3\cdot(1)^2\cdot(-1)^3(-1.01+1)\]which is the answer.
For the second part, we only need to calculate the difference of $L(0.98,-1.01)$ and $(0.98)^2(-1.01)^3$, and then divide the difference by the actual value $(0.98)^2(-1.01)^3$, and multiply $100$ to get percentage. In other words, the following is the answer \(\frac{L(0.98,-1.01)-(0.98)^2(-1.01)^3}{(0.98)^2(-1.01)^3}\cdot 100\\%\)