<s>[INST] Determine if function $f(x)=\ln \left ( \frac{x+1}{x-1} \right )$ is odd or even. Determine if function $f(x)=\ln \left ( \frac{x+1}{x-1} \right )$ is odd or even.
My solution:
$$
\begin{align}
f(-x)&=\ln\left ( \frac{-x+1}{-x-1} \right )\\
&=\ln \left ( \frac{-(x-1)}{-(x+1)}\right )\\
&=\ln\left ( \frac{x-1}{x+1} \right )\\
&=\ln (x-1)-\ln (x+1)\\
&=-(\ln (x+1)-\ln (x-1))\\
&=-\ln \left ( \frac{x+1}{x-1} \right ) = -f(x)
\end{align}
$$
It seems that function is odd. However, according to WolframAlpha it's neither odd nor even. 
 [/INST] While your conclusion is fine, there are still some problems with your computations, where some cautions is needed.
By definition, an even or odd function must have a symmetric domain of definition, so the first thing is to make this domain explicit: $\mathcal{D} := (-\infty, 1) \cup (1, +\infty)$.
Then we want to prove that $f$ is odd. In other words, we have to show that, for $x \in \mathcal{D}$, 
$$f(-x) = -f(x).$$
If $x >1$, we have $x+1>0$ and $x-1>0$, so we can indeed write $f(x) = \ln (x+1)-\ln(x-1)$. However, for $x<1$, we have $x+1<0$ and $x-1<0$, so the expression $\ln (x+1)-\ln(x-1)$ is meaningless. Hence, the computation you gave is only valid for $x>1$. This is enough to prove your claim, but you need to make this explicit.
Another solution is to avoid splitting the logarithm altogether. For any $x\in\mathcal{D}$,
$$f(-x) = \ln \left(\frac{x-1}{x+1} \right) = \ln \left(\frac{1}{\left(\frac{x+1}{x-1}\right)} \right) = -\ln \left(\frac{x+1}{x-1} \right) = -f(x).$$
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