So lets see how to resolve inequalities that a bit longer than usual.

$\frac{x+3}{x-1} \textless x \textless \frac{1}{2}$

$\frac{x+3}{x-1} \textless x$ and $x \textless \frac{1}{2}$

$\frac{x+3}{x-1} -x \textless 0$

$\frac{x+3-x(x-1)}{x-1} \textless 0$

$\frac{x+3-x^2+x}{x-1} \textless 0$

$\frac{-x^2+2x+3}{x-1} \textless 0$

$\frac{(-x+3)(x+1)}{x-1} \textless 0$

$\Rightarrow -1 \textless x \textless 1$ or $x >3$ and $x \textless \frac{1}{2}$.

Thus, $-1 \textless x \textless \frac{1}{2}$