The gradient of the curve is given by \(\frac{dy}{dx} = k \frac{y}{\sqrt{x+1}}\), where \(k\) is a constant of proportionality.

Separate variables: \(\frac{1}{y} dy = \frac{k}{\sqrt{x+1}} dx\).

Integrate both sides: \(\int \frac{1}{y} dy = \int \frac{k}{\sqrt{x+1}} dx\).

This gives \(\ln y = 2k \sqrt{x+1} + c\), where \(c\) is the constant of integration.

Using the point \((0, 1)\), substitute into the equation: \(\ln 1 = 2k \sqrt{0+1} + c\), which simplifies to \(c = -2k\).

Using the point \((3, e)\), substitute into the equation: \(\ln e = 2k \sqrt{3+1} - 2k\), which simplifies to \(1 = 4k - 2k\).

Solve for \(k\): \(k = \frac{1}{2}\).

Substitute \(k = \frac{1}{2}\) and \(c = -1\) back into the equation: \(\ln y = \sqrt{x+1} - 1\).

Exponentiate both sides to solve for \(y\): \(y = \exp(\sqrt{x+1} - 1)\).