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. 

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 $x\in\mathcal{D}$,
$$f(-x) = \ln \left(\frac{x-1}{x+1} \right) = \ln \left(\frac{1}{\left(\frac{x+1}{x-1}\right)} \right) = -\ln \left(\frac{x+1}{x-1} \right) = -f(x).$$
