(a) Start by separating variables: \(e^{-y} dy = x e^{-x} dx\).

Integrate both sides: \(\int e^{-y} dy = \int x e^{-x} dx\).

The left side integrates to \(-e^{-y}\).

For the right side, use integration by parts: let \(u = x\) and \(dv = e^{-x} dx\), then \(du = dx\) and \(v = -e^{-x}\).

Integration by parts gives: \(\int x e^{-x} dx = -xe^{-x} - \int -e^{-x} dx = -xe^{-x} - e^{-x} + C\).

Thus, \(-e^{-y} = -xe^{-x} - e^{-x} + C\).

Using the initial condition \(y = 0\) when \(x = 0\), we find \(-1 = -1 + C\), so \(C = 0\).

Therefore, \(-e^{-y} = -xe^{-x} - e^{-x}\).

Solve for \(y\): \(e^{-y} = (x+1)e^{-x}\), so \(y = -\ln((x+1)e^{-x})\).

(b) Substitute \(x = 1\) into the expression for \(y\):

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

Thus, \(y = 1 - \ln 2\).