(a) Separate variables: \(\dfrac{dy}{y(y+1)}=-x^{-2}\,dx\). With \(1/[y(y+1)]=1/y-1/(y+1)\), integration gives \(\ln|y/(y+1)|=1/x+C\). At \((1,1)\), \(C=-1-\ln2\).

The branch containing the initial point has \(y/(y+1)=\tfrac12e^{1/x-1}\). Solving gives \(\boxed{y=\dfrac{e^{1/x-1}}{2-e^{1/x-1}}}\).

(b) As \(x\to\infty\), \(e^{1/x-1}\to e^{-1}\), so \(\boxed{y\to1/(2e-1)}\).