(a) Differentiate the equation \(x^3 + y^3 + 2xy + 8 = 0\) implicitly with respect to \(x\):

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

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

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

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

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

(b) Find the gradients at the points \((0, -2)\) and \((-2, 0)\):

At \((0, -2)\), substitute into \(\frac{dy}{dx} = \frac{-3(0)^2 - 2(-2)}{3(-2)^2 + 2(0)} = \frac{4}{12} = \frac{1}{3}\).

At \((-2, 0)\), substitute into \(\frac{dy}{dx} = \frac{-3(-2)^2 - 2(0)}{3(0)^2 + 2(-2)} = \frac{-12}{-4} = 3\).

Use the formula for the tangent of the angle between two lines: \(\tan \alpha = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|\), where \(m_1 = \frac{1}{3}\) and \(m_2 = 3\).

\(\tan \alpha = \left| \frac{\frac{1}{3} - 3}{1 + \frac{1}{3} \times 3} \right| = \left| \frac{-\frac{8}{3}}{2} \right| = \frac{4}{3}\).