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

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

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

2. Apply the discriminant condition for real roots.

For $y$ to be in the range, there must be at least one real value of $x$ that corresponds to it. Hence, the discriminant of the quadratic equation in $x$ must be non-negative ($\Delta \ge 0$).
$(9y-3)^2 - 4(18)(3y-1) \ge 0$

3. Simplify the discriminant inequality.

Factorising to simplify the expansion:
$9(3y-1)^2 - 72(3y-1) \ge 0$
$9(3y-1)[(3y-1) - 8] \ge 0$
$9(3y-1)(3y-9) \ge 0$
$27(3y-1)(y-3) \ge 0$
$(3y-1)(y-3) \ge 0$

4. Solve the quadratic inequality to find the range.

The critical values are $y = \frac{1}{3}$ and $y = 3$. Since the quadratic expression represents a 'U-shaped' parabola and is $\ge 0$, we take the regions outside the roots:
$y \le \frac{1}{3} \text{ or } y \ge 3$

1. Find the equations of the asymptotes by rewriting the rational function.

$\begin{aligned} y &= \frac{-18x^2+3x+1}{9x+3} \\ &= \frac{-2x(9x+3) + 9x+1}{9x+3} \\ &= -2x + \frac{1(9x+3)-2}{9x+3} \\ &= -2x + 1 - \frac{2}{9x+3} \end{aligned}$
The vertical asymptote is $x = -\frac{1}{3}$ and the oblique asymptote is $y = -2x + 1$.

2. Find the exact coordinates of the stationary points using the range of $y$ found in part (a).

From part (a), the range is $y \le \frac{1}{3}$ or $y \ge 3$. The boundary values give the $y$-coordinates of the stationary points.
Using the rearranged equation from part (a): $18x^2 + (9y - 3)x + (3y - 1) = 0$
Substitute $y = \frac{1}{3}$:
$18x^2 = 0 \implies x = 0$
Substitute $y = 3$:
$18x^2 + 24x + 8 = 0 \implies 2(3x+2)^2 = 0 \implies x = -\frac{2}{3}$
The stationary points are $(0, \frac{1}{3})$ and $(-\frac{2}{3}, 3)$.

3. Determine axial intercepts and produce the sketch.

$x$-intercepts occur when $y = 0$:
$1+3x-18x^2 = 0 \implies (1-3x)(1+6x) = 0 \implies x = \frac{1}{3} \text{ or } -\frac{1}{6}$.
$y$-intercept is at $x=0 \implies y=\frac{1}{3}$ (which coincides with the local maximum point).
The sketch should clearly display two branches of a hyperbola confined by the asymptotes $x=-\frac{1}{3}$ and $y=-2x+1$, with the right branch having a maximum at $(0, \frac{1}{3})$ and the left branch having a minimum at $(-\frac{2}{3}, 3)$.

1. Identify the relationship between the new equation and curve $C$, and state the first transformation.

Let the equation of curve $C$ be $y = \mathrm{f}(x) = \frac{1+3x-18x^{2}}{9x+3}$.
The new equation is obtained by replacing every $x$ with $2x - 1$, so it is given by $y = \mathrm{f}(2x - 1)$.
The first transformation is a translation of $1$ unit in the positive $x$-direction, which transforms $y = \mathrm{f}(x)$ to $y = \mathrm{f}(x - 1)$.

2. State the second transformation to achieve the final equation.

The second transformation is a scaling parallel to the $x$-axis by a scale factor of $\frac{1}{2}$.
This replaces $x$ with $2x$, transforming $y = \mathrm{f}(x - 1)$ to $y = \mathrm{f}(2x - 1)$.