To find the coordinates of \(M\), we start by differentiating both sides of the equation \((x^2 + y^2)^2 = 2(x^2 - y^2)\) with respect to \(x\).

The left-hand side is \((x^2 + y^2)^2\). Using the chain rule, the derivative is:

\(2(x^2 + y^2) \cdot (2x + 2y \frac{dy}{dx})\).

The right-hand side is \(2(x^2 - y^2)\). Its derivative is:

\(4x - 4y \frac{dy}{dx}\).

Setting the derivatives equal gives:

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

Rearranging terms to solve for \(\frac{dy}{dx}\), we set \(\frac{dy}{dx} = 0\) to find horizontal tangents:

\(2(x^2 + y^2) \cdot 2x = 4x\).

Simplifying gives:

\((x^2 + y^2) = 1\).

Substituting \(x^2 + y^2 = 1\) into the original equation:

\((1)^2 = 2(x^2 - y^2)\).

This simplifies to:

\(1 = 2x^2 - 2y^2\).

Using \(x^2 + y^2 = 1\), we solve for \(x^2\) and \(y^2\):

\(x^2 = \frac{1}{2} + y^2\).

Substitute into \(1 = 2x^2 - 2y^2\):

\(1 = 2(\frac{1}{2} + y^2) - 2y^2\).

\(1 = 1 + 2y^2 - 2y^2\).

\(x^2 = \frac{3}{4}\), \(y^2 = \frac{1}{4}\).

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

(a) Differentiate the equation \(3x^2 + 4xy + 3y^2 = 5\) implicitly with respect to \(x\):

\(\frac{d}{dx}(3x^2) = 6x\),

\(\frac{d}{dx}(4xy) = 4x\frac{dy}{dx} + 4y\),

\(\frac{d}{dx}(3y^2) = 6y\frac{dy}{dx}\).

Set the derivative equal to zero:

\(6x + 4x\frac{dy}{dx} + 4y + 6y\frac{dy}{dx} = 0\).

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

\(4x\frac{dy}{dx} + 6y\frac{dy}{dx} = -6x - 4y\),

\(\frac{dy}{dx}(4x + 6y) = -6x - 4y\),

\(\frac{dy}{dx} = -\frac{6x + 4y}{4x + 6y}\),

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

(b) The slope of the tangent is \(-2\) (parallel to \(y + 2x = 0\)).

Set \(\frac{dy}{dx} = -2\):

\(-\frac{3x + 2y}{2x + 3y} = -2\),

\(3x + 2y = 4x + 6y\),

\(x = -4y\) or \(y = -\frac{x}{4}\).

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

\(3(-4y)^2 + 4(-4y)y + 3y^2 = 5\),

\(48y^2 - 16y^2 + 3y^2 = 5\),

\(35y^2 = 5\),

\(y^2 = \frac{1}{7}\),

\(y = \pm \frac{1}{\sqrt{7}}\).

For \(y = \frac{1}{\sqrt{7}}\), \(x = -4\left(\frac{1}{\sqrt{7}}\right) = -\frac{4}{\sqrt{7}}\).

For \(y = -\frac{1}{\sqrt{7}}\), \(x = 4\left(\frac{1}{\sqrt{7}}\right) = \frac{4}{\sqrt{7}}\).

The coordinates are \(\left( \frac{4}{\sqrt{7}}, \frac{1}{\sqrt{7}} \right)\) and \(\left( -\frac{4}{\sqrt{7}}, -\frac{1}{\sqrt{7}} \right)\).

To find the gradient of the curve, we need to differentiate the given equation implicitly with respect to \(x\).

The equation is \(3e^{2x}y + e^xy^3 = 14\).

Differentiate each term:

1. Differentiate \(3e^{2x}y\) using the product rule:

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

2. Differentiate \(e^xy^3\) using the product rule:

\(\frac{d}{dx}(e^xy^3) = e^xy^3 \cdot \frac{dy}{dx} + 3e^xy^2\).

Combine these results to get the derivative of the left-hand side:

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

Substitute \(x = 0\) and \(y = 2\) into the equation:

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

Simplify:

\(12 + 3 \frac{dy}{dx} + 8 \frac{dy}{dx} + 12 = 0\).

Combine like terms:

\(24 + 11 \frac{dy}{dx} = 0\).

Solve for \(\frac{dy}{dx}\):

\(11 \frac{dy}{dx} = -24\).

\(\frac{dy}{dx} = -\frac{24}{11}\).

Thus, the gradient of the curve at the point \((0, 2)\) is \(-\frac{4}{3}\).

To find the \(x\)-coordinate of the maximum point \(M\), we need to differentiate the given equation implicitly with respect to \(x\).

The given equation is \(x^3 + xy^2 + ay^2 - 3ax^2 = 0\).

Differentiating implicitly, we have:

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

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

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

Set \(\frac{dy}{dx} = 0\) for the maximum point:

\(6ax - 3x^2 - y^2 = 0\).

Substitute \(y^2 = \frac{3ax^2 - x^3}{x + a}\) into the equation:

\(6ax - 3x^2 - \frac{3ax^2 - x^3}{x + a} = 0\).

Simplify and solve for \(x\):

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

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

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

Factor out \(x\):

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

Since \(x = 0\) is not a maximum point, solve:

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

Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 3\), \(b = -3a\), \(c = 6a^2\):

\(x = \frac{3a \pm \sqrt{(-3a)^2 - 4 \cdot 3 \cdot 6a^2}}{6}\).

\(x = \frac{3a \pm \sqrt{9a^2 - 72a^2}}{6}\).

\(x = \frac{3a \pm \sqrt{-63a^2}}{6}\).

\(x = \frac{3a \pm i\sqrt{63}a}{6}\).

Since we need a real solution, consider the simplified form:

\(x = \sqrt{3a}\).

(i) To find the gradient where the curve crosses the y-axis, substitute \(x = 0\) into the equation. The y-intercept is \(y = \frac{1 + 0^2}{1 + e^{0}} = \frac{1}{2}\).

Differentiate \(y = \frac{1 + x^2}{1 + e^{2x}}\) using the quotient rule:

\(\frac{d}{dx} \left( \frac{1 + x^2}{1 + e^{2x}} \right) = \frac{(1 + e^{2x}) \cdot 2x - (1 + x^2) \cdot 2e^{2x}}{(1 + e^{2x})^2}\).

Substitute \(x = 0\) into the derivative:

\(\frac{(1 + e^{0}) \cdot 0 - (1 + 0^2) \cdot 2e^{0}}{(1 + e^{0})^2} = \frac{-2}{4} = -\frac{1}{2}\).

(ii) For \(2x^3 + 5xy + y^3 = 8\), substitute \(x = 0\) to find \(y\):

\(2(0)^3 + 5(0)y + y^3 = 8 \Rightarrow y^3 = 8 \Rightarrow y = 2\).

Differentiate implicitly:

\(\frac{d}{dx}(2x^3) = 6x^2\), \(\frac{d}{dx}(5xy) = 5y + 5x \frac{dy}{dx}\), \(\frac{d}{dx}(y^3) = 3y^2 \frac{dy}{dx}\).

Combine to get:

\(6x^2 + 5y + 5x \frac{dy}{dx} + 3y^2 \frac{dy}{dx} = 0\).

Substitute \(x = 0\), \(y = 2\):

\(0 + 5(2) + 0 \cdot \frac{dy}{dx} + 3(2)^2 \frac{dy}{dx} = 0 \Rightarrow 10 + 12 \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = -\frac{5}{6}\).