+9665 \(e^{g\left(x\right)}=\frac{x^x}{x^2-1}\)
\(g(x)= \ln(\frac{x^x}{x^2-1})\\ \quad \;\;\;=x\ln x-(\ln(x+1)(x-1))\\ \quad \;\;\;=x \ln x - \ln (x+1)- \ln (x-1)\\ \text{Sorry if I am wrong I am pretty bad at logarithms}\)
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+118659
HI MAX 
Your answer looks good except if we are dealing with the real number system there are restrictions on the domain.
\(e^{g\left(x\right)}=\frac{x^x}{x^2-1}\\ \mbox{e to the power of any real number is positive so the RHS must be positive}\\ g(x)= \ln(\frac{x^x}{x^2-1})\\ \)
1) \(x^2-1\ne0\;\;\;\;so\\ x\ne\pm1\)
2) xx does crazy things isf x is not not greater than zero. (x certainly cannot equal 0) So I am going to say that x must be greater than 0.
I think that this is correct but I would like another mathematician to comment
\(x>0 \qquad x\ne1\)
3)
\(\frac{x^x}{x^2-1}>0\\ \mbox{Since the numerator is positive the denominator must be positive too}\\ x^2>1\\ x>1 \;\;x<-1\\ so\;\;x>1\)
\(g(x)= \ln(\frac{x^x}{x^2-1})\\ \quad \;\;\;=x\ln x-(\ln(x+1)(x-1))\\ \quad \;\;\;=x \ln x - \ln (x+1)- \ln (x-1)\\ \)
Your answer is correct so long as you state that the domain is \(x>1 \;\;for \;\;x\in R\)
