You can differentiate your function by using the chain rule, the quotient rule, and the power rule.

First, you need to recognize that your function can be expressed as

#f(u) = u^(1/3)#, where

#u = (x-1)/(x+3)#

The chain rule allows you to differentiate a function #y# that depends on a variable #u#, which in turn depends on a variable #x# by using the formula

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

The derivative of #y# will thus be equal to

#d/dx(y) = d/(du)(u^(1/3)) * d/dx((x-1)/(x-3))#

#d/dx(y) = 1/3u^(1/3 - 1) * d/dx((x-1)/(x+3))#

To determine #d/dx((x-1)/(x+3))#, you can use the quotient rule, which allows you to differentiate functions that can be expressed as a *quotient* of two other functions

#a(x) = (b(x))/(c(x))#

by using the formula

#color(blue)(d/dx(a(x)) = (b^'(x) * c(x) - b(x) * c^'(x))/[c(x)]^2#, with #c(x)!=0#.

In your case, #d/dx((x-1)/(x+3))# can be differentiated by using the quotient rule to get

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

#d/dx((x-1)/(x+3)) = (color(red)(cancel(color(black)(x))) + 3 - color(red)(cancel(color(black)(x))) + 1)/(x+3)^2#

#d/dx((x-1)/(x+3)) = 4/(x+3)^2#

Plug this into the derivative of #y# to get

#d/dx(y) = 1/3 * ((x-1)/(x+3))^(-2/3) * 4/(x+3)^2#

#d/dx(y) = 4/3 * 1/( (x-1)^(2/3) * 1/(x+3)^(2/3) * (x+3)^2)#

#d/dx(y) = 4/3 * 1/((x-1)^(2/3) * (x+3)^(4/3)) = color(green)(4/3 * (x-1)^(-2/3) * (x+3)^(-4/3))#

**Alternatively**, you can complicate things a bit by writing your original function as

#y = (x-1)^(1/3)/(x+3)^(1/3)#

and using a combination of the power rule and quotient rule.

This will get *a little* messy, but you could write

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

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

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

This is equivalent to

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

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

#d/dx(y) = 1/3 * [(x-1)^(1/3) * (x + 3)^(-1/3) * (color(red)(cancel(color(black)(x))) + 3 - color(red)(cancel(color(black)(x))) + 1)/((x-1)(x+3))]#

Finally, you have

#d/dx(y) = color(green)(4/3 * (x-1)^(-2/3) * (x+3)^(-4/3))#