Answer
$(-\infty,-5)\cup (-1,1)$
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Work Step by Step
$\mathrm{See\:the\:graph\:above.}$
$a.\:\:\:$ By observing the graph, we come to know that the point of intersection is $\:x=-5.\:$ Based on the graph, the function $\:y=\frac{3}{x-1}\:$ (represented by red curve) is below the function $\:\frac{2}{x+1}\:$ (represented by blue curve), when:
$x < -5\:\:$ and $\:\:-1< x <1\:\:$ (vertical asymptotes are at $\:x=1\:\:\:\mathrm{and}\:\:x=-1\:$ because for those numbers the denominators of the functions are zero.)
$b.$
$\frac{3}{x-1} -5$
$x^2-1 < 0\:\:\rightarrow x^2 < 1\:\:$ when $\:\:-1 < x < 1$
Solution of this case is $\:x\in(-1,1).$
$\mathrm{Case\:II:}$
$x+5 < 0,\:\:$ when $\:\:x < -5$
$x^2-1 > 0\:\:\rightarrow\:\: x^2 > 1\:\:$ when $\:\:-1 > x > 1$
Solution of this case is $\:x\in(-\infty,-5).$
Combined solution from both cases: $\:(-\infty,-5)\cup (-1,1).$
