To find the gradient of the curve, we need to differentiate \(y = \frac{1+x}{1+2x}\) with respect to \(x\).

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

First, find \(\frac{du}{dx} = 1\) and \(\frac{dv}{dx} = 2\).

Applying the quotient rule:

\(\frac{dy}{dx} = \frac{(1+2x)(1) - (1+x)(2)}{(1+2x)^2}\)

Simplify the numerator:

\((1+2x) - 2(1+x) = 1 + 2x - 2 - 2x = -1\)

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

Since \((1+2x)^2 > 0\) for \(x > -\frac{1}{2}\), \(\frac{-1}{(1+2x)^2} < 0\).

Therefore, the gradient of the curve is always negative for \(x > -\frac{1}{2}\).