Answer
$$\eqalign{
& f\left( {x,y} \right) = {x^2}y \cr
& dz = 0.44 \cr} $$
Work Step by Step
$$\eqalign{
& {\left( {2.01} \right)^2}\left( {9.02} \right) - {2^2} \cdot 9 \cr
& {\text{Rewrite the expression}} \cr
& = {\left( {2 + 0.01} \right)^2}\left( {9 + 0.02} \right) - {2^2} \cdot 9 \cr
& z = \underbrace {{{\left( {2 + 0.01} \right)}^2}}_{{{\left( {x + \Delta x} \right)}^2}}\underbrace {\left( {9 + 0.02} \right)}_{\left( {y + \Delta y} \right)} - \underbrace {{2^2} \cdot 9}_{{x^2}y} \cr
& {\text{Let the function }} \cr
& z = f\left( {x,y} \right) = {x^2}y,{\text{ with }}x = 2{\text{ and }}y = 9,{\text{ }} \cr
& \Delta x = 0.01{\text{ and }}\Delta y = 0.02 \cr
& {\text{Therefore, the total differential is }} \cr
& dz = \frac{\partial }{{\partial x}}\left[ {{x^2}y} \right]dx + \frac{\partial }{{\partial y}}\left[ {{x^2}y} \right]dy \cr
& dz = 2xydx + {x^2}dy \cr
& {\text{With }}dx = \Delta x{\text{ and }}dy = \Delta y \cr
& dz = 2\left( 2 \right)\left( 9 \right)\left( {0.01} \right) + {\left( 2 \right)^2}\left( {0.02} \right) \cr
& dz = 0.44 \cr
& \cr
& f\left( {x,y} \right) = {x^2}y \cr
& dz = 0.44 \cr} $$