Answer
$(-\infty,0)\cup(1,\infty)$
or $\{x|x \lt 0$ or $x \gt 1\}$
Work Step by Step
First, we see that if we want $\displaystyle \frac{x}{x-1}$ to be defined, we must have
$x\neq 1\quad $
Next,
the argument of a logarithmic function must be greater than zero.
So, for h to be defined, we must have
$\displaystyle \frac{x}{x-1} \gt 0 \quad $
... $x-1$ changes signs at $1$, x changes signs at 0
$Q(x)=\displaystyle \frac{x}{x-1}$ is zero or undefined for $x=0,1$.
We test intervals into which these two numbers split the number line.
$\left[\begin{array}{llll}
\text{interval} & (-\infty,0) & (0,1) & (1,\infty)\\
\text{test point } t & -1 & 0.5 & 2\\
Q(t) & \frac{-1}{-1-1} & \frac{0.5}{0.5-1} & \frac{2}{2-1}\\
\text{sign of }Q(t) & + & - & +
\end{array}\right]$
Domain: $(-\infty,0)\cup(1,\infty)$
or $\{x|x \lt 0$ or $x \gt 1\}$