First of all, let's simplify the fraction to #3/(8x^3)#.

Two options:

**Using the Quotient Rule**

In case you didn't know, if #y = f(x)/g(x)#, then

#dy/dx = (f'(x)g(x) - g'(x)f(x))/(g(x)^2)#

In this case, #f(x) = 3# and #g(x) = 8x^3#. Therefore,

#dy/dx = ([3]' * (8x^3) - [8x^3]' * 3)/(8x^3)^2#.

#= (0 * (8x^3) - [8x^3]' * 3)/(8x^3)^2# (because #[3]' = 0#)

#= (-[8x^3]' * 3)/(8x^3)^2#

#= (-24x^2 * 3)/(8x^3)^2# (because #[8x^3]' = 24x^2#)

#= (-72x^2)/(64x^6)# (simplifying)

#= (-9x^2)/(8x^6)# (reducing)

#= -9/(8x^4)# (reducing exponents)

That is one way of doing it, but it's kind of hard. There is a simpler way of doing it.

**Using the Power Rule**

#y = 3/(8x)^3 = 3/8x^(-3)#

In case you didn't know, if #y = x^n#, then #dy/dx = n * x^(n-1)#. Therefore,

#dy/dx = 3/8 * (-3) * x^(-3-1)#

#= -9/8 * x^(-4)#

#= -9/(8x^4)#