Answer
Solution set = $\{3\}.$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{llllll}
x-1 \gt 0, & and & x+3 \gt 0 & and & x \gt 0 & \\
x\gt 1 & & x\gt -3 & & x\gt 0 &
\end{array}\right]$
$ x \gt 1 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\displaystyle \log_{2}(x-1)-\log_{2}(x+3)=\log_{2}(\frac{1}{x})$ ... LHS: quotient rule
$\displaystyle \log_{2}(\frac{x-1}{x+3})=\log_{2}(\frac{1}{x})$ ... $\log_{2}$ is one-to-one
$\displaystyle \frac{x-1}{x+3}=\frac{1}{x}$
Multiply by $x(x+3)$, the LCD:
$ x(x-1)=x+3$
$ x^{2}-x=x+3$
$ x^{2}-2x-3=0$
Factor: +1 and -3 are factors of -3 whose sum is -2.
$(x+1)(x-3)=0$
$ x=3 \qquad $ ... satisfies (*)
$ x=-1 \quad $ ... does not satisfy (*)
Thus, the solution set = $\{3\}.$
