To find the coordinates where the gradient of the tangent is \(\frac{1}{4}\), we first need to find the derivative of \(y = \frac{x}{1 + \ln x}\).

Using the quotient rule, \(\frac{d}{dx} \left( \frac{u}{v} \right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}\), where \(u = x\) and \(v = 1 + \ln x\).

Then, \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = \frac{1}{x}\).

So, \(\frac{dy}{dx} = \frac{(1 + \ln x) \cdot 1 - x \cdot \frac{1}{x}}{(1 + \ln x)^2} = \frac{1 + \ln x - 1}{(1 + \ln x)^2} = \frac{\ln x}{(1 + \ln x)^2}\).

Set \(\frac{dy}{dx} = \frac{1}{4}\):

\(\frac{\ln x}{(1 + \ln x)^2} = \frac{1}{4}\).

Cross-multiply to get:

\(4 \ln x = (1 + \ln x)^2\).

Expand and rearrange to form a quadratic equation:

\((\ln x)^2 - 2 \ln x + 1 = 0\).

Solving this quadratic equation, we find \(\ln x = 1\), so \(x = e\).

Substitute \(x = e\) back into the original equation to find \(y\):

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

Thus, the exact coordinates are \((e, \frac{1}{2}e)\).