Answer
$\dfrac{y-1}{y+1}=cx^2$
Work Step by Step
Need to separate the variables to determine the differential equation and then integrate the obtained equation on the both sides.
$\int \dfrac{1}{y^2-1}dy=\int \dfrac{dx}{x}$
This implies that
$\dfrac{\ln [{\dfrac{y-1}{y+1}}]}{2}=\ln |x|+c'$
or, $\ln [{\dfrac{y-1}{y+1}}]=2 \ln |x|+\ln c'$
Hence, $\dfrac{y-1}{y+1}=cx^2$
