First, separate the variables:

\(\frac{1}{y^2} dy = 4xe^{-2x} dx\).

Integrate both sides:

\(\int \frac{1}{y^2} dy = \int 4xe^{-2x} dx\).

The left side integrates to:

\(-\frac{1}{y}\).

For the right side, use integration by parts:

Let \(u = x\) and \(dv = 4e^{-2x} dx\).

Then \(du = dx\) and \(v = -2e^{-2x}\).

So, \(\int 4xe^{-2x} dx = -2xe^{-2x} - \int -2e^{-2x} dx\).

\(= -2xe^{-2x} + e^{-2x}\).

Thus, the integrated equation is:

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

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

\(-1 = -2(0)e^{0} + e^{0} + C\).

\(-1 = 1 + C\).

\(C = -2\).

Substitute back:

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

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

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

Multiply numerator and denominator by \(e^{2x}\):

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