(a) Differentiate the equation \(x^3 + y^2 + 3x^2 + 3y = 4\) implicitly with respect to \(x\):

\(3x^2 + 2y \frac{dy}{dx} + 6x + 3 \frac{dy}{dx} = 0\).

Rearrange to solve for \(\frac{dy}{dx}\):

\(2y \frac{dy}{dx} + 3 \frac{dy}{dx} = -3x^2 - 6x\).

\(\frac{dy}{dx} (2y + 3) = -3x^2 - 6x\).

\(\frac{dy}{dx} = -\frac{3x^2 + 6x}{2y + 3}\).

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

Set the numerator of \(\frac{dy}{dx} = -\frac{3x^2 + 6x}{2y + 3}\) to zero:

\(3x^2 + 6x = 0\).

Factorize:

\(3x(x + 2) = 0\).

\(x = 0\) or \(x = -2\).

Substitute \(x = 0\) into the original equation:

\(y^2 + 3y - 4 = 0\).

Factorize:

\((y - 1)(y + 4) = 0\).

\(y = 1\) or \(y = -4\).

Coordinates: \((0, 1)\) and \((0, -4)\).

Substitute \(x = -2\) into the original equation:

\(y^2 + 3y = 0\).

Factorize:

\(y(y + 3) = 0\).

\(y = 0\) or \(y = -3\).

Coordinates: \((-2, 0)\) and \((-2, -3)\).