Notice that your function can be written in the form

#y = e^(1/2x)/(2x^3) = f(x)/g(x)#,

which means that you can use the quotient rule to find its derivative.

The derivative of a function that can be expressed as the quotient of two other functions can be found by using

#color(blue)(d/dx(y) = (f^'(x) * g(x) - f(x) * g^'(x))/[g(x)]^2)#, where #g(x) !=0#.

This means that you can write

#y^' = (d/dx(e^(1/2x)) * (2x^3) - e^(1/2x) * d/dx(2x^3))/(2x^3)^2#

To find #d/dx(e^(1/2x))# you can use the chain rule, which helps you differentiate functions that depends on a variable #u#, which in turn depend on a variable #x#.

#color(blue)(d/dx(h(u)) = d/(du)y * d/dx(u))#

In your case, you have #h(u) = e^u# and #u = 1/2x#, so that

#d/x(h(u)) = d/(du)e^u * d/dx(1/2x)#

#d/dx(h(u)) = e^u * 1/2 = 1/2 * e^(1/2x)#

Plug this into your target derivative to get

#y^' = (1/color(red)(cancel(color(black)(2))) * e^(1/2x) * color(red)(cancel(color(black)(2)))x^3 - e^(1/2x) * 6x^2)/(4x^6)#

#y^' = (x^3 * e^(1/2x) - 6x^2 * e^(1/2x))/(4x^6)#

This can be further simplified to get

#y^' = (color(red)(cancel(color(black)(x^2))) * e^(1/2x) * (x - 6))/(4x^(color(red)(cancel(color(black)(6))) color(blue)(4))) = color(green)((e^(1/2x)(x-6))/(4x^4)#

**Alternatively**, you could rewrite your original function as

#y = 1/2 * e^(1/2x) * x^(-3)#

and use the product, chain, and power rules to get

#y^' = 1/2 * [d/dx(e^(1/2x)) * x^(_3) + e^(1/2x) * d/dx(x^(-3))]#

#y^' = 1/2 * [1/2 * e^(1/2x) * x^(-3) + e^(1/2x) * (-3)x^(-4)]#

#y^' = 1/2 * e^(1/2x)(1/(2x^3) -3/x^4)#

which will get you

#y^' = 1/2 * e^(1/2x) * ((x - 6))/(2x^4) = color(green)((e^(1/2x) * (x-6))/(4x^4))#