Ответ:

Производная дроби равна  [tex]\bf \Big(\dfrac{u}{v}\Big)'=\dfrac{u'v-uv'}{v^2}[/tex]  ,    

[tex]\bf (\sqrt[3]{u})'=(u^{1/3})'=\dfrac{1}{3}\cdot u^{\frac{1}{3}-1}\cdot u'\ \ ,\ \ \ (e^{^{u}})'=e^{^{u}}\cdot u'[/tex]  

  [tex]y=\dfrac{\sqrt[3]{2x^2-3x+1}}{e^{^{-x}}}\\\\\\y'=\dfrac{\dfrac{1}{3}\cdot (2x^2-3x+1)^{-\frac{2}{3}}\cdot (4x-3)\cdot e^{^{-x}}-\sqrt[3]{2x^2-3x+1}\cdot (-e^{^{-x}})}{e^{^{-2x}}}=\\\\\\=\dfrac{e^{^{-x}}\cdot \Big(\dfrac{1}{3}\cdot (2x^2-3x+1)^{-\frac{2}{3}}\cdot (4x-3)+\sqrt[3]{2x^2-3x+1}\Big)}{e^{^{-2x}}}=\\\\\\=\dfrac{\dfrac{1}{3}\cdot (2x^2-3x+1)^{-\frac{2}{3}}\cdot (4x-3)+\sqrt[3]{2x^2-3x+1}}{e^{^{-x}}}=\\\\\\=e^{^{x}}\cdot \Big(\dfrac{4x-3}{3\cdot \sqrt[3]{(2x^2-3x+1)^2}}+ \sqrt[3]{2x^2-3x+1}\Big)=[/tex]

[tex]=e^{^{x}}\cdot \Big(\dfrac{4x-3+3\cdot (2x^2-3x+1)}{3\, \sqrt[3]{(2x^2-3x+1)^2}}\Big)=\dfrac{e^{^{x}}\cdot (6x^2-5x)}{3\, \sqrt[3]{(2x^2-3x+1)^2}}=\dfrac{x\cdot e^{^{x}}\cdot (6x-5)}{3\, \sqrt[3]{(2x^2-3x+1)^2}}[/tex]