(a) Start by separating variables:

\(\frac{1}{y^2 + y} dy = -\frac{1}{x^2} dx\)

Express \(\frac{1}{y^2 + y}\) in partial fractions:

\(\frac{1}{y(y+1)} = \frac{1}{y} - \frac{1}{y+1}\)

Integrate both sides:

\(\int \left(\frac{1}{y} - \frac{1}{y+1}\right) dy = \int -\frac{1}{x^2} dx\)

\(\ln |y| - \ln |y+1| = \frac{1}{x} + C\)

Use the initial condition \(x = 1\), \(y = 1\) to find \(C\):

\(\ln 1 - \ln 2 = \frac{1}{1} + C\)

\(-\ln 2 = 1 + C\)

\(C = -1 - \ln 2\)

Substitute back to find \(y\):

\(\ln \frac{y}{y+1} = \frac{1}{x} - 1 - \ln 2\)

\(\frac{y}{y+1} = e^{\frac{1}{x} - 1 - \ln 2}\)

\(\frac{y}{y+1} = \frac{e^{\frac{1}{x} - 1}}{2}\)

\(y = \frac{e^{\frac{1}{x} - 1}}{2 - e^{\frac{1}{x} - 1}}\)

(b) As \(x\) tends to infinity, \(\frac{1}{x}\) tends to 0, so:

\(y = \frac{e^{0 - 1}}{2 - e^{0 - 1}} = \frac{e^{-1}}{2 - e^{-1}} = \frac{1}{2e - 1}\)