(a) Differentiate both sides of \(\ln(x+y) = x - 2y\) with respect to \(x\).

The derivative of the left-hand side using the chain rule is \(\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right)\).

The derivative of the right-hand side is \(1 - 2\frac{dy}{dx}\).

Equate the derivatives: \(\frac{1}{x+y} \left(1 + \frac{dy}{dx}\right) = 1 - 2\frac{dy}{dx}\).

Multiply through by \(x+y\): \(1 + \frac{dy}{dx} = (x+y)(1 - 2\frac{dy}{dx})\).

Expand and rearrange: \(1 + \frac{dy}{dx} = x+y - 2(x+y)\frac{dy}{dx}\).

Combine terms: \(1 + \frac{dy}{dx} + 2(x+y)\frac{dy}{dx} = x+y\).

Factor out \(\frac{dy}{dx}\): \(\frac{dy}{dx}(1 + 2(x+y)) = x+y - 1\).

Solve for \(\frac{dy}{dx}\): \(\frac{dy}{dx} = \frac{x+y-1}{2(x+y)+1}\).

(b) For the tangent to be parallel to the \(x\)-axis, \(\frac{dy}{dx} = 0\).

Set \(\frac{x+y-1}{2(x+y)+1} = 0\), which implies \(x+y-1 = 0\).

Thus, \(x+y = 1\).

Substitute \(x+y = 1\) into the original equation: \(\ln(1) = x - 2y\).

Since \(\ln(1) = 0\), we have \(x - 2y = 0\).

Solving \(x+y = 1\) and \(x - 2y = 0\) simultaneously:

From \(x - 2y = 0\), \(x = 2y\).

Substitute into \(x+y = 1\): \(2y + y = 1\).

Thus, \(3y = 1\) and \(y = \frac{1}{3}\).

Substitute \(y = \frac{1}{3}\) into \(x = 2y\): \(x = \frac{2}{3}\).

The coordinates are \(x = \frac{2}{3}, y = \frac{1}{3}\).