(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 \(m_2 = 3\).

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

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

(a) Differentiate \(ye^{2x}\) with respect to \(x\):

\(\frac{d}{dx}(ye^{2x}) = 2ye^{2x} + e^{2x} \frac{dy}{dx}\).

Differentiate \(y^2 e^x\) with respect to \(x\):

\(\frac{d}{dx}(y^2 e^x) = 2y e^x \frac{dy}{dx} + y^2 e^x\).

Set the derivative of the left-hand side to zero and solve for \(\frac{dy}{dx}\):

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

Rearrange to find \(\frac{dy}{dx}\):

\(\frac{dy}{dx} = \frac{2ye^x - y^2}{2y - e^x}\).

(b) For the tangent to be parallel to the y-axis, \(\frac{dy}{dx}\) is undefined, which occurs when the denominator is zero:

\(2y - e^x = 0 \Rightarrow y = \frac{e^x}{2}\).

Substitute \(y = \frac{e^x}{2}\) into the original equation:

\(\left(\frac{e^x}{2}\right)e^{2x} - \left(\frac{e^x}{2}\right)^2 e^x = 2\).

Simplify to find \(x\):

\(\frac{e^{3x}}{2} - \frac{e^{3x}}{4} = 2 \Rightarrow \frac{e^{3x}}{4} = 2 \Rightarrow e^{3x} = 8\).

\(3x = \ln 8 \Rightarrow x = \frac{1}{3} \ln 8 = \ln 2\).

Substitute back to find \(y\):

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

The coordinates are \((\ln 2, 1)\).

(a) Differentiate the equation \(x^3 + 3xy^2 - y^3 = 5\) implicitly with respect to \(x\):

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

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

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

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

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

(b) For the tangent to be parallel to the y-axis, \(\frac{dy}{dx}\) is undefined, which occurs when the denominator is zero:

\(y^2 - 2xy = 0\).

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

So, \(y = 0\) or \(y = 2x\).

For \(y = 0\), substitute into the original equation:

\(x^3 = 5\), so \(x = \sqrt[3]{5}\).

Point: \(\left( \sqrt[3]{5}, 0 \right)\).

For \(y = 2x\), substitute into the original equation:

\(x^3 + 3x(2x)^2 - (2x)^3 = 5\).

\(x^3 + 12x^3 - 8x^3 = 5\).

\(5x^3 = 5\), so \(x = 1\).

Then \(y = 2 \times 1 = 2\).

Point: (1, 2).

To find where the tangent is parallel to the \(x\)-axis, we need \(\frac{dy}{dx} = 0\).

Differentiate the given equation implicitly:

\(\frac{d}{dx}(2x^2y - xy^2) = \frac{d}{dx}(a^3)\).

Using the product rule, we have:

\(4xy + 2x^2\frac{dy}{dx} - (y^2 + 2xy\frac{dy}{dx}) = 0\).

Simplify to get:

\(4xy + 2x^2\frac{dy}{dx} - y^2 - 2xy\frac{dy}{dx} = 0\).

Combine terms:

\((2x^2 - 2xy)\frac{dy}{dx} = y^2 - 4xy\).

Set \(\frac{dy}{dx} = 0\) to find:

\(y^2 - 4xy = 0\).

Factor to get:

\(y(y - 4x) = 0\).

So, \(y = 0\) or \(y = 4x\).

Reject \(y = 0\) because it does not satisfy the original equation unless \(a = 0\), which is not allowed as \(a\) is positive.

Substitute \(y = 4x\) into the original equation:

\(2x^2(4x) - x(4x)^2 = a^3\).

Simplify to get:

\(8x^3 - 16x^3 = a^3\).

\(-8x^3 = a^3\).

\(x^3 = -\frac{a^3}{8}\).

\(x = -\frac{a}{2}\).

Substitute \(x = -\frac{a}{2}\) back into \(y = 4x\):

\(y = 4\left(-\frac{a}{2}\right) = -2a\).

Thus, the \(y\)-coordinate of the point is \(y = -2a\).