The curve $C$ has equation given by $y = \dfrac{1+3x-18x^2}{9x+3}$.

1. Rearrange the equation of the curve into a quadratic equation in terms of $x$.

$y = \frac{1+3x-18x^2}{9x+3}$
$y(9x+3) = 1+3x-18x^2$
$9xy + 3y = 1 + 3x - 18x^2$
$18x^2 + (9y-3)x + (3y-1) = 0$

2. Apply the condition for real roots to form an inequality in $y$.

For $x$ to be real, the discriminant of the quadratic equation must be non-negative ($\Delta \ge 0$):
$b^2 - 4ac \ge 0$
$(9y-3)^2 - 4(18)(3y-1) \ge 0$

3. Simplify the inequality without a calculator by extracting common factors.

$9(3y-1)^2 - 72(3y-1) \ge 0$
Factorising out the common term $9(3y-1)$:
$9(3y-1)[(3y-1) - 8] \ge 0$
$9(3y-1)(3y-9) \ge 0$
$27(3y-1)(y-3) \ge 0$
Dividing by 27:
$(3y-1)(y-3) \ge 0$

4. Solve the quadratic inequality to determine the valid range of $y$.

Solving $(3y-1)(y-3) \ge 0$, the values of $y$ lie outside the roots $\frac{1}{3}$ and $3$:
$y \le \frac{1}{3} \text{ or } y \ge 3$

1. Find the equations of the asymptotes using algebraic long division.

$y = \frac{1+3x-18x^2}{9x+3}$
By performing algebraic long division, we can rewrite the rational function as:
$y = -2x + 1 - \frac{2}{9x+3}$
From the denominator, the vertical asymptote is $9x + 3 = 0 \implies x = -\frac{1}{3}$.
From the linear quotient, the oblique (slant) asymptote is $y = -2x + 1$.

2. Determine the coordinates of the stationary points and axial intercepts.

Using a graphing calculator (or by setting $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$), we locate the stationary points. Using the range derived in part (a), the turning points occur at $y = \frac{1}{3}$ and $y = 3$.
The local maximum point is $(0, \frac{1}{3})$.
The local minimum point is $(-\frac{2}{3}, 3)$.
To find the $x$-intercepts, set $y = 0$:
$1+3x-18x^2 = 0$
$(1+6x)(1-3x) = 0 \implies x = -\frac{1}{6} \text{ or } x = \frac{1}{3}$

3. Sketch the graph showing all key features.

Draw the asymptotes $x = -\frac{1}{3}$ and $y = -2x + 1$ with dotted lines.
Sketch the upper branch containing the minimum point $(-\frac{2}{3}, 3)$, staying bounded above $y = -2x + 1$ as $x \to -\infty$.
Sketch the lower branch containing the maximum point $(0, \frac{1}{3})$ (which is also the $y$-intercept), passing through the $x$-intercepts $(-\frac{1}{6}, 0)$ and $(\frac{1}{3}, 0)$, and bounded below $y = -2x + 1$ as $x \to \infty$.

1. Express the new equation in terms of the original function.

Let $f(x) = \dfrac{1+3x-18x^2}{9x+3}$ be the equation of curve $C$.

The target equation is $\dfrac{1+3(2x-1)-18(2x-1)^2}{9(2x-1)+3}$.
This is formed by replacing every $x$ in $f(x)$ with $(2x-1)$. Hence, the new equation is $y = f(2x-1)$.

2. State the sequence of transformations that maps $f(x)$ to $f(2x-1)$.

We can build $f(2x-1)$ through the following sequence:

1. Replace $x$ with $x-1$: This corresponds to a translation of $1$ unit in the positive $x$-direction, transforming $y = f(x)$ to $y = f(x-1)$.
2. Replace $x$ with $2x$: This corresponds to a scaling parallel to the $x$-axis by a scale factor of $\frac{1}{2}$, transforming $y = f(x-1)$ to $y = f(2x-1)$.

(Note: An alternative valid sequence is a scaling parallel to the $x$-axis by a scale factor of $\frac{1}{2}$ to get $f(2x)$, followed by a translation of $\frac{1}{2}$ units in the positive $x$-direction to get $f(2(x-\frac{1}{2})) = f(2x-1)$.)