The Folium of Descartes is a curve, defined by the equation $x^{3} + y^{3} = 3axy$, where $a$ is a real constant. It is given that $a \ne 0$ for this question.

1. Differentiate both sides of the equation implicitly with respect to $x$.

$\frac{d}{dx}(x^3 + y^3) = \frac{d}{dx}(3axy)$
$3x^2 + 3y^2 \frac{dy}{dx} = 3a \left( y + x\frac{dy}{dx} \right)$
$3x^2 + 3y^2 \frac{dy}{dx} = 3ay + 3ax \frac{dy}{dx}$

2. Isolate the derivative term $\dfrac{dy}{dx}$ to show the required expression.

$3y^2 \frac{dy}{dx} - 3ax \frac{dy}{dx} = 3ay - 3x^2$
$3\frac{dy}{dx}(y^2 - ax) = 3(ay - x^2)$
$\frac{dy}{dx} = \frac{ay - x^2}{y^2 - ax}$

1. Evaluate the gradient of the tangent at the point $\left(\frac{3a}{2},\frac{3a}{2}\right)$ using the derivative from part (a).

$\frac{dy}{dx} = \frac{a\left(\frac{3a}{2}\right) - \left(\frac{3a}{2}\right)^2}{\left(\frac{3a}{2}\right)^2 - a\left(\frac{3a}{2}\right)}$

$\frac{dy}{dx} = \frac{\frac{3a^2}{2} - \frac{9a^2}{4}}{\frac{9a^2}{4} - \frac{3a^2}{2}} = \frac{-\frac{3a^2}{4}}{\frac{3a^2}{4}} = -1$

2. Calculate the gradient of the normal to the curve.

Let the gradient of the normal be $m_n$.
$m_n = -\frac{1}{\frac{dy}{dx}} = -\frac{1}{-1} = 1$

3. Find the equation of the normal and show it is independent of $a$.

Using the point-slope form with $m_n = 1$ and the point $\left(\frac{3a}{2}, \frac{3a}{2}\right)$:
$y - \frac{3a}{2} = 1 \left(x - \frac{3a}{2}\right)$
$y = x$
Since the equation of the normal simplifies to $y = x$, it does not contain the constant $a$ and is therefore independent of the value of $a$.

1. Set $\frac{dy}{dx} = 0$ to find a relationship between $p$ and $q$.

At a stationary point, $\frac{dy}{dx} = 0$.
Using the expression for $\frac{dy}{dx}$ from part (a):
$\frac{ay - x^2}{y^2 - ax} = 0 \implies ay - x^2 = 0 \implies x^2 = ay$
Substitute the coordinates of the stationary point $(pa, qa)$:
$(pa)^2 = a(qa)$
$p^2 a^2 = q a^2$
Since $a \neq 0$, we can divide by $a^2$:
$q = p^2$

2. Substitute the point into the curve equation to find a second relationship.

The point $(pa, qa)$ lies on the curve, so it must satisfy the equation $x^3 + y^3 = 3axy$:
$(pa)^3 + (qa)^3 = 3a(pa)(qa)$
$p^3 a^3 + q^3 a^3 = 3pqa^3$
Since $a \neq 0$, dividing the entire equation by $a^3$ gives:
$p^3 + q^3 = 3pq$

3. Solve the simultaneous equations for $p$.

Substitute $q = p^2$ into $p^3 + q^3 = 3pq$:
$p^3 + (p^2)^3 = 3p(p^2)$
$p^3 + p^6 = 3p^3$
$p^6 - 2p^3 = 0$
$p^3(p^3 - 2) = 0$
Since $p$ is a positive constant, $p \neq 0$. Thus:
$p^3 = 2 \implies p = 2^{\frac{1}{3}} \text{ (or } \sqrt[3]{2} \text{)}$

4. Solve for $q$.

Substitute $p = 2^{\frac{1}{3}}$ back into $q = p^2$:
$q = \left(2^{\frac{1}{3}}\right)^2 = 2^{\frac{2}{3}} \text{ (or } \sqrt[3]{4} \text{)}$