The equation of a curve is $y=\dfrac{x^2-3a^2}{x-a}$, where $a$ is a positive constant.

1. Rewrite the function to find the equations of the asymptotes.

$y = \frac{x^2-3a^2}{x-a}$
$y = \frac{x^2-a^2-2a^2}{x-a}$
$y = \frac{(x-a)(x+a)-2a^2}{x-a}$
$y = x + a - \frac{2a^2}{x-a}$
From this form, the vertical asymptote is obtained when the denominator is zero, so $x = a$.
As $x \to \pm\infty$, $\frac{2a^2}{x-a} \to 0$, so the oblique asymptote is $y = x + a$.

2. Differentiate the function with respect to $x$.

Using the rewritten form $y = x + a - 2a^2(x-a)^{-1}$:
$\frac{dy}{dx} = 1 - 2a^2(-1)(x-a)^{-2}$
$\frac{dy}{dx} = 1 + \frac{2a^2}{(x-a)^2}$

3. Show that the derivative cannot be zero to prove there are no stationary points.

For a stationary point to exist, we require $\frac{dy}{dx} = 0$.
Since $a$ is a positive constant, $a > 0$, implying $2a^2 > 0$.
For all $x$ in the domain ($x \neq a$), the squared term $(x-a)^2 > 0$.
Therefore, the fraction $\frac{2a^2}{(x-a)^2} > 0$, which gives:
$\frac{dy}{dx} = 1 + \frac{2a^2}{(x-a)^2} > 1$
Since $\frac{dy}{dx} > 0$ for all $x \neq a$, $\frac{dy}{dx}$ can never be $0$.
Thus, there are no stationary points on the curve.

1. Find the coordinates of the points of intersection with the axes.

For the $y$-intercept, set $x = 0$:
$y = \frac{0 - 3a^2}{0 - a} = 3a$
So the $y$-intercept is $(0, 3a)$.

For the $x$-intercepts, set $y = 0$:
$\frac{x^2 - 3a^2}{x - a} = 0$
$x^2 - 3a^2 = 0$
$x = \pm \sqrt{3}a$
So the $x$-intercepts are $(\sqrt{3}a, 0)$ and $(-\sqrt{3}a, 0)$.

2. Determine the behaviour of the curve relative to its asymptotes and sketch the graph.

From part (a), the asymptotes are $x = a$ and $y = x + a$, and the curve has no stationary points (it is strictly increasing on its domain).

Since $a$ is a positive constant, $\sqrt{3}a > a$.

• Left branch ($x < a$): Passes through the $x$-intercept $(-\sqrt{3}a, 0)$ and the $y$-intercept $(0, 3a)$. It approaches the oblique asymptote $y = x + a$ from above as $x \to -\infty$, and approaches the vertical asymptote $x = a$ upwards ($y \to \infty$) as $x \to a^-$.
• Right branch ($x > a$): Passes through the $x$-intercept $(\sqrt{3}a, 0)$. It emerges from the vertical asymptote $x = a$ downwards ($y \to -\infty$) as $x \to a^+$, and approaches the oblique asymptote $y = x + a$ from below as $x \to \infty$.

(A properly scaled sketch showing these two continuous branches, with the asymptotes drawn as dashed lines and the axes intercepts clearly labelled, should be presented.)

1. Find the expression for $f'(x)$.

By algebraic long division or rewriting the numerator:
$f(x) = \frac{x^2-a^2-2a^2}{x-a} = x+a-\frac{2a^2}{x-a}$
$f'(x) = 1 - 2a^2(-1)(x-a)^{-2} = 1 + \frac{2a^2}{(x-a)^2}$

2. Determine the range of $x$ for which $f'$ is increasing.

For $f'$ to be increasing, its derivative $f''(x)$ must be positive.
$f''(x) = -4a^2(x-a)^{-3} = \frac{-4a^2}{(x-a)^3}$
Since $a$ is a positive constant, $-4a^2 < 0$.
For $f''(x) > 0$, we require the denominator to be negative:
$(x-a)^3 < 0$
$x - a < 0$
$x < a$

(Alternatively, by considering the graph of $y = 1 + \frac{2a^2}{(x-a)^2}$, there is a vertical asymptote at $x = a$. As $x$ approaches $a$ from the left ($x < a$), the positive denominator $(x-a)^2$ gets closer to 0, causing $f'(x)$ to increase towards $+\infty$. Hence $f'$ is increasing for $x < a$.)