Answer
$$\Delta y = 3{x^2}\Delta x + 3x\Delta {x^2} + \Delta {x^3}\,\,\,{\text{ and}}\,\,\,{\text{ }}dy = 3{x^2}dx$$
Work Step by Step
$$\eqalign{
& {\text{Let }}y = {x^3} \cr
& \cr
& {\text{Calculating }}\Delta y.\,\,\,f\left( x \right) = {x^3} \cr
& \Delta y = f\left( {x + \Delta x} \right) - f\left( x \right) \cr
& {\text{Then evaluating }} \cr
& \Delta y = {\left( {x + \Delta x} \right)^3} - {x^3} \cr
& \Delta y = {x^3} + 3{x^2}\Delta x + 3x\Delta {x^2} + \Delta {x^3} - {x^3} \cr
& {\text{Simplifying}} \cr
& \Delta y = 3{x^2}\Delta x + 3x\Delta {x^2} + \Delta {x^3} \cr
& \cr
& {\text{and}} \cr
& y = {x^3} \cr
& \frac{{dy}}{{dx}} = 3{x^2},{\text{ so }}dy = 3{x^2}dx \cr
& \cr
& \Delta y = 3{x^2}\Delta x + 3x\Delta {x^2} + \Delta {x^3}\,\,\,{\text{ and}}\,\,\,{\text{ }}dy = 3{x^2}dx \cr} $$
