Separate variables: \[\int\frac1y\,dy=\int\frac1{(x+1)(3x+1)}\,dx.\] Resolve into partial fractions: \[\frac1{(x+1)(3x+1)}=-\frac1{2(x+1)}+\frac3{2(3x+1)}.\] Thus \[\ln|y|=-\frac12\ln|x+1|+\frac12\ln|3x+1|+C.\] Using \((x,y)=(1,1)\) gives \(C=-\tfrac12\ln2\). On the interval containing 1, the initial positive solution is \[\boxed{y=\sqrt{\frac{3x+1}{2(x+1)}}}\quad(x>-1/3).\] At \(x=3\), \[y=\sqrt{10/8}=\boxed{\frac{\sqrt5}2}.\]