1. The gradient of the curve is given by \(\frac{dy}{dx} = k \frac{y^2}{x}\), where \(k\) is a constant.

2. Separate variables: \(\frac{1}{y^2} dy = \frac{k}{x} dx\).

3. Integrate both sides: \(-\frac{1}{y} = k \ln x + C\), where \(C\) is the constant of integration.

4. Use the point \((1, 1)\) to find \(C\): \(-1 = k \ln 1 + C \Rightarrow C = -1\).

5. Use the point \((e, 2)\) to find \(k\): \(-\frac{1}{2} = k \ln e - 1 \Rightarrow k = \frac{1}{2}\).

6. Substitute \(k\) and \(C\) back into the integrated equation: \(-\frac{1}{y} = \frac{1}{2} \ln x - 1\).

7. Solve for \(y\): \(y = \frac{2}{2 - \ln x}\).