\(\frac{3}{x-1}+\frac{4}{x+1}=\frac{3x+2}{1-x^2}\Leftrightarrow\frac{3}{x-1}+\frac{4}{x+1}=-\frac{3x+2}{x^2-1}\Leftrightarrow\frac{3}{x-1}+\frac{4}{x+1}=-\frac{3x+2}{\left(x-1\right)\left(x+1\right)}\)\(\Leftrightarrow3.\left(x+1\right)+4.\left(x-1\right)=-\left(3x+2\right)\)
=> 3x + 3 + 4x - 4 + 3x + 2 = 0
=> 10x + 1 = 0
=> x = -1/10