A curve C has equation $x^3 + y^3 - xy = A$ where $A$ is a non-zero constant.

1. Differentiate the equation implicitly with respect to $x$.

Differentiating both sides of $x^3 + y^3 - xy = A$ with respect to $x$:
$3x^2 + 3y^2 \frac{dy}{dx} - \left(1 \cdot y + x \frac{dy}{dx}\right) = 0$

2. Use the condition for stationary points to find the relationship between $x$ and $y$.

At a stationary point, the gradient of the curve is zero, so $\frac{dy}{dx} = 0$. Substituting this into the differentiated equation:
$3x^2 + 3y^2(0) - \left(y + x(0)\right) = 0$
$3x^2 - y = 0$
$y = 3x^2 \quad \text{(Shown)}$

1. Substitute the equation of the locus of stationary points into the curve's equation to form a polynomial in $x$.

Substituting $y = 3x^2$ into $x^3 + y^3 - xy = A$:
$x^3 + (3x^2)^3 - x(3x^2) = A$
$x^3 + 27x^6 - 3x^3 = A$
$27x^6 - 2x^3 - A = 0$

2. Solve the resulting quadratic equation in $x^3$ using the quadratic formula.

Let $u = x^3$. The equation is $27u^2 - 2u - A = 0$.
Using the quadratic formula:
$u = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(27)(-A)}}{2(27)}$
$x^3 = \frac{2 \pm \sqrt{4 + 108A}}{54}$
$x^3 = \frac{1 \pm \sqrt{1 + 27A}}{27}$

3. Express the $x$-coordinates of the stationary points in terms of $A$.

Taking the cube root of both sides:
$x = \sqrt[3]{\frac{1 \pm \sqrt{1 + 27A}}{27}}$
$x = \frac{1}{3} \sqrt[3]{1 \pm \sqrt{1 + 27A}}$

4. Apply the discriminant condition for the quadratic equation in $x^3$ to find the range of $A$ for two distinct stationary points.

For the curve to have two distinct stationary points, the quadratic equation in $x^3$ must have two distinct real roots. Thus, its discriminant must be strictly positive:
$(-2)^2 - 4(27)(-A) > 0$
$4 + 108A > 0$
$1 + 27A > 0 \implies A > -\frac{1}{27}$
Since the problem states $A$ is a non-zero constant, $A \neq 0$.
The range of values is $A > -\frac{1}{27}, A \neq 0$.

1. Recalculate the $x$-coordinates of the stationary points (from part (b)) to use in evaluating their nature.

Substitute $y = 3x^2$ into the curve equation $x^3 + y^3 - xy = A$:
$x^3 + (3x^2)^3 - x(3x^2) = A$
$27(x^3)^2 - 2x^3 - A = 0$
Using the quadratic formula to solve for $x^3$:
$x^3 = \frac{2 \pm \sqrt{(-2)^2 - 4(27)(-A)}}{54} = \frac{1 \pm \sqrt{1 + 27A}}{27}$
So the $x$-coordinates are $x = \left( \frac{1 \pm \sqrt{1 + 27A}}{27} \right)^{\frac{1}{3}}$.

2. Find the second derivative $\frac{d^2y}{dx^2}$ evaluated at the stationary points.

From part (a), we have the first derivative implicitly defined by the equation:
$3x^2 + 3y^2 \frac{dy}{dx} - y - x\frac{dy}{dx} = 0$
Differentiating this implicitly with respect to $x$ gives:
$6x + 6y\left(\frac{dy}{dx}\right)^2 + 3y^2 \frac{d^2y}{dx^2} - \frac{dy}{dx} - \left( \frac{dy}{dx} + x\frac{d^2y}{dx^2} \right) = 0$
At a stationary point, $\frac{dy}{dx} = 0$. Substituting this in yields:
$6x + 3y^2 \frac{d^2y}{dx^2} - x\frac{d^2y}{dx^2} = 0$
$\frac{d^2y}{dx^2}(3y^2 - x) = -6x \implies \frac{d^2y}{dx^2} = \frac{-6x}{3y^2 - x}$

3. Express the second derivative entirely in terms of $x$.

Substitute the locus of stationary points, $y = 3x^2$, into the expression for $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{-6x}{3(3x^2)^2 - x} = \frac{-6x}{27x^4 - x}$
Factoring out $x$ (since $A \neq 0$, the origin $(0,0)$ does not lie on the curve, so $x \neq 0$):
$\frac{d^2y}{dx^2} = \frac{-6x}{x(27x^3 - 1)} = \frac{-6}{27x^3 - 1}$

4. Evaluate the sign of $\frac{d^2y}{dx^2}$ for each stationary point to determine its nature.

Using the expressions for $x^3$, we evaluate the denominator $27x^3 - 1$:
$27x^3 - 1 = 27\left( \frac{1 \pm \sqrt{1 + 27A}}{27} \right) - 1 = \pm \sqrt{1 + 27A}$

For the stationary point where $x = \left( \frac{1 + \sqrt{1 + 27A}}{27} \right)^{\frac{1}{3}}$:
$\frac{d^2y}{dx^2} = \frac{-6}{+\sqrt{1 + 27A}} < 0$
Hence, this is a local maximum.

For the stationary point where $x = \left( \frac{1 - \sqrt{1 + 27A}}{27} \right)^{\frac{1}{3}}$:
$\frac{d^2y}{dx^2} = \frac{-6}{-\sqrt{1 + 27A}} = \frac{6}{\sqrt{1 + 27A}} > 0$
Hence, this is a local minimum.