First, separate the variables:

\(\frac{dy}{y^3} = ke^{-x} dx\)

Integrate both sides:

\(\int y^{-3} \, dy = \int ke^{-x} \, dx\)

This gives:

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

Using the initial condition \(y = 1\) when \(x = 0\):

\(-\frac{1}{2(1)^2} = -k + c\)

\(-\frac{1}{2} = -k + c\)

Using the second condition \(y = \sqrt{e}\) when \(x = 1\):

\(-\frac{1}{2(\sqrt{e})^2} = -ke^{-1} + c\)

\(-\frac{1}{2e} = -\frac{k}{e} + c\)

Solving these equations gives \(k = \frac{1}{2}\) and \(c = 0\).

Substitute back to find \(y\):

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

\(y^2 = e^x\)

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