Helpppp

Find the derivative of the following via implicit differentiation:
d/dx(y) = d/dx((log(((1 + x)/(-1 + x))^log(3)))/log(3))

Using the chain rule, d/dx(y) = ( dy(u))/( du) ( du)/( dx), where u = x and d/( du)(y(u)) = y'(u):
d/dx(x) y'(x) = d/dx((log(((1 + x)/(-1 + x))^log(3)))/log(3))

The derivative of x is 1:
1 y'(x) = d/dx((log(((1 + x)/(-1 + x))^log(3)))/log(3))

Factor out constants:
y'(x) = (d/dx(log(((1 + x)/(-1 + x))^log(3))))/(log(3))

Simplify log(((x + 1)/(x - 1))^log(3)) using the identity log(a^b) = b log(a):
y'(x) = (d/dx(log(3) log((1 + x)/(-1 + x))))/log(3)

Factor out constants:
y'(x) = (log(3) d/dx(log((1 + x)/(-1 + x))))/log(3)

Simplify the expression:
y'(x) = d/dx(log((1 + x)/(-1 + x)))

Simplify log((x + 1)/(x - 1)) using the identity log(a/b) = log(a) - log(b):
y'(x) = d/dx(log(1 + x) - log(-1 + x))

Differentiate the sum term by term and factor out constants:
y'(x) = d/dx(log(1 + x)) - d/dx(log(-1 + x))

Using the chain rule, d/dx(log(x - 1)) = ( dlog(u))/( du) ( du)/( dx), where u = x - 1 and d/( du)(log(u)) = 1/u:
y'(x) = d/dx(log(1 + x)) - (d/dx(-1 + x))/(x - 1)

Differentiate the sum term by term:
y'(x) = d/dx(log(1 + x)) - (d/dx(-1) + d/dx(x))/(x - 1)

The derivative of -1 is zero:
y'(x) = d/dx(log(1 + x)) - (d/dx(x) + 0)/(-1 + x)

Simplify the expression:
y'(x) = -(d/dx(x))/(-1 + x) + d/dx(log(1 + x))

The derivative of x is 1:
y'(x) = d/dx(log(1 + x)) - 1/(x - 1)

Using the chain rule, d/dx(log(x + 1)) = ( dlog(u))/( du) ( du)/( dx), where u = x + 1 and d/( du)(log(u)) = 1/u:
y'(x) = -1/(-1 + x) + (d/dx(1 + x))/(x + 1)

Differentiate the sum term by term:
y'(x) = -1/(-1 + x) + (d/dx(1) + d/dx(x))/(x + 1)

The derivative of 1 is zero:
y'(x) = -1/(-1 + x) + (d/dx(x) + 0)/(1 + x)

Simplify the expression:
y'(x) = -1/(-1 + x) + (d/dx(x))/(1 + x)

The derivative of x is 1: 
 y'(x) = -1/(-1 + x) + 1/(x + 1)  - Courtesy of Mathematica 11 Home Edition

Could also do it as follows:

And yet another way without substitutions

y = log ₃   [ (x +1) / (x - 1) ] ^ln3    by a log property

y = ln 3  log₃  [ (x + 1) / ( x - 1)]

Since ln 3  is  just  constant...it won't affect the differentiation.....we can go back and add it at the end

y  =  log ₃  [ (x + 1) / ( x - 1) ]

y  = log ₃ (x + 1)  - log₃ (x - 1) 

Note :  log _b  a  =   ln a / ln b

y  =  ln  (x + 1) / ln 3  -  ln (x - 1) / ln 3

y  =  (1/ ln3)  [ ln ( x + 1) - ln ( x - 1) ]

y' = (1/ln3)  [ 1/(x + 1) - 1 / (x - 1] ]

Adding  ln 3  back we have

y'  =  ln3  ( 1/ ln 3)   ( 1/ (x + 1) - 1/ (x - 1) ]

y ' =  1/ (x + 1)  - 1 / (x - 1)