Находим производную:
y'(x)=((3x^2-4x+2)'(x^2+2x+3)+(x^2+2x+3)'(3x^2-4x+2))/(x^2+2x+3)^2=
((6x^3+8x^2+18x-4x^2-8x-12)+(6x^3-8x^2+4x+6x^2-8x+4))/(x^2+2x+3)^2=
(6x^3+4x^2+10x-12+6x^3-2x^2-4x+4)/(x^2+2x+3)^2=
(12x^3+10x^2+6x-8)/(x^2+2x+3)^2
При x=1,y'(1)=(12*1+10*1+6*1-8)/(1+2*2+3)^2=
20/8=5
Ответ:5