You're dealing with a product of **three functions**

#y(u) = f(u) * g(u) * h(u)#

which means that the formula for differentiating using the product rule will take the form

#color(blue)(d/(du)(y) = [d/(du)(f(u))] * g(u) * h(u) + f(u) * [d/(du)(g(u))] * h(u) + f(u) * g(u) * d/(du)(h(u)))#

In your case, you have

#f(u) = u^(-2) + u^(-3)#

#g(u) = u^5 - 2u^2#

#h(u) = e^(2u)#

The derivative of #y(u)# will thus be

#d/(du)(y(u)) = [d/(du)(u^(-2) + u^(-3))] * (u^5 - 2u^2) * e^(2u) + (u^(-2) + u^(-3)) * [d/(du)(u^5 - 2u^2)] * e^(2u) + (u^(-2) + u^(-3)) * (u^5 - 2u^2) * d/dx(e^(2u))#

To make the calculations easier to follow, I'll take each of those three derivatives separately.

#d/(du)(u^(-2) + u^(-3)) = d/(du)u^(-2) + d/(du)u^(-3)#

#d/(du)(u^(-2) + u^(-3)) = -2u^(-3) + (-3u^(-4))#

#d/(du)(u^(-2) + u^(-3)) = -2u^(-3) - 3u^(-4)#

Next, you have

#d/(du)(u^5 - 2u^2) = d/(du)u^5 - 2d/(du)u^2#

#d/(du)(u^5 - 2u^2) = 5u^4 - 4u#

To differentiate #e^(2u)#, use the chain rule for #e^t#, with #t=2u#. This will get you

#d/(du)(e^t) = d/(dt)(e^t) * d/(du)(t)#

#d/(du)(e^t) = e^t * 2#

#d/(du)(e^(2u)) = 2e^(2u)#

I'll break up the target derivative further to write the three products separately. The first product is

#[d/(du)(u^-2 + u^-3)] * (u^5 - 2u^2) * e^(2u)#

#(-2u^(-3) - 3u^-4) * (u^5 - 2u^2) * e^(2u)#

#(-2u^2 + 4u^-1 - 3u + 6u^-2) * e^(2u) -> color(blue)((P_1))#

The second product is

#(u^-2 + u^-3) * [d/(du)(u^5 - 2u^2)] * e^(2u)#

#(u^-2 + u^-3) * (5u^4 - 4u) * e^(2u)#

#(5u^2 - 4u^-1 + 5u - 4u^-2) * e^(2u) -> color(blue)((P_2))#

The third product is

#(u^-2 + u^-3) * (u^5 - 2u^2) * d/(du)e^(2u)#

#2 * (u^3 - 2 + u^-2 - 2u^-1) * e^(2u) -> color(blue)((P_3))#

The target derivative will thus be

#y^' = color(blue)(P_1) + color(blue)(P_2) + color(blue)(P_3)#

#y^' = e^(2u) * ( -2u^-2 + 4u^-1 - 3u + 6u^-2 + 5u^2 - 4u^-1 + 5u - 4u^-2 + 2u^3 - 4 +2u^-2 - 4u^-1)#

This can be simplified to

#y^' = color(green)(e^(2u) * (2u^3 + 5u^2 + 2u -4 - 4u^(-1) + 2u^(-2)))#