Answer
$$\,\,\,\,\,dy = 21$$
Work Step by Step
$$\eqalign{
& y = 4{x^3} - 3x;\,\,\,\,\,\,x = 3{\text{ and }}\Delta x = 0.2 \cr
& {\text{The differential }}dy{\text{ is the product of }}f'\left( x \right){\text{ and }}dx,{\text{ or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)dx \cr
& {\text{Find }}f'\left( x \right) \cr
& f'\left( x \right) = \left( {4{x^3} - 3x} \right)' \cr
& f'\left( x \right) = 12{x^2} - 3 \cr
& {\text{Then}}{\text{, the differential }}dy{\text{ is}} \cr
& \,\,\,\,\,dy = \left( {12{x^2} - 3} \right)dx \cr
& {\text{As }}dx{\text{ approaches 0}}{\text{, the value of }}dy{\text{ gets closer and closer that of }}\Delta y,{\text{ }} \cr
& {\text{so that for small nonzero values of }}dx, \cr
& \,\,\,\,\,dy \approx \Delta y, \cr
& {\text{or}} \cr
& \,\,\,\,\,dy = f'\left( x \right)\Delta x \cr
& {\text{Then }} \cr
& \,\,\,\,\,dy = \left( {12{x^2} - 3} \right)\Delta x \cr
& {\text{Substituting }}x = 3{\text{ and }}\Delta x = dx = 0.2 \cr
& \,\,\,\,\,dy = \left( {12{{\left( 3 \right)}^2} - 3} \right)\left( {0.2} \right) \cr
& {\text{simplifying}} \cr
& \,\,\,\,\,dy = 21 \cr} $$
