Answer
$f\left( {2.01,1.02} \right) = {\left( {2.01} \right)^3}{\left( {1.02} \right)^2} \simeq 8.44$
Using a calculator:
${\left( {2.01} \right)^3}{\left( {1.02} \right)^2} \simeq 8.44867$.
Work Step by Step
Let $f\left( {x,y} \right) = {x^3}{y^2}$. So, the partial derivatives are
${f_x} = 3{x^2}{y^2}$, ${\ \ \ }$ ${f_y} = 2{x^3}y$
We can consider ${\left( {2.01} \right)^3}{\left( {1.02} \right)^2}$ as a value of $f\left( {x,y} \right) = {x^3}{y^2}$. Thus,
$f\left( {2.01,1.02} \right) = {\left( {2.01} \right)^3}{\left( {1.02} \right)^2}$
Using the linear approximation, Eq. (3) we have
$f\left( {a + h,b + k} \right) \approx f\left( {a,b} \right) + {f_x}\left( {a,b} \right)h + {f_y}\left( {a,b} \right)k$
For $\left( {a,b} \right) = \left( {2,1} \right)$ and $\left( {h,k} \right) = \left( {0.01,0.02} \right)$, we get
$f\left( {2.01,1.02} \right) \approx f\left( {2,1} \right) + {f_x}\left( {2,1} \right)\cdot0.01 + {f_y}\left( {2,1} \right)\cdot0.02$
$f\left( {2.01,1.02} \right) \simeq 8 + 12\cdot0.01 + 16\cdot0.02$
$f\left( {2.01,1.02} \right) = {\left( {2.01} \right)^3}{\left( {1.02} \right)^2} \simeq 8.44$
Using a calculator, we get ${\left( {2.01} \right)^3}{\left( {1.02} \right)^2} \simeq 8.44867$.
