Your answer is CORRECT.

Find the general solution of $x{y}^{\text{'}}+3y=4x^{-2}{e}^{-4x}$

$y^{\text{'}}+\frac{3}{x}y=\frac{4e^{-4}}{x^3}$

a$)y=4x^{-3}{e}^{-4x}+C{x}^{-3}$

b$)y=-x^{-3}{e}^{-4x}+C{x}^{-3}$

c$)y=\frac{1}{4}{x}^{-3}{e}^{-4x}+C{x}^{-3}$

$(d)$

$y(x^3)=\frac{4}{-4}$

d$)y=-x^{-3}{e}^{4x}+C{x}^{-3}$

$y=\frac{-e}{x^3}$