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

1. Find the point(s) of intersection with the coordinate axes.

Substitute $x = 0$ into the equation of $C$:
$y = \frac{0^2 - 3(0) + 3}{1 - 0} = 3$
The $y$-intercept is $(0, 3)$.

Substitute $y = 0$:
$x^2 - 3x + 3 = 0$
The discriminant $\Delta = (-3)^2 - 4(1)(3) = 9 - 12 = -3$.
Since $\Delta < 0$, there are no real roots, hence the curve does not intersect the $x$-axis.

2. Determine the equations of the asymptotes.

The denominator cannot be zero, giving the vertical asymptote:
$1 - x = 0 \implies x = 1$

To find the oblique asymptote, perform polynomial division on $\frac{x^2 - 3x + 3}{1-x}$:
$\frac{x^2 - 3x + 3}{-x + 1} = \frac{-x(1-x) - 2x + 3}{1-x} = -x + \frac{-2(1-x) + 1}{1-x} = -x + 2 + \frac{1}{1-x}$
As $x \to \pm\infty$, $\frac{1}{1-x} \to 0$, so the oblique asymptote is:
$y = -x + 2$

3. Find the coordinates of the turning points.

Differentiate $y = -x + 2 + (1-x)^{-1}$ with respect to $x$:
$\frac{dy}{dx} = -1 + (1-x)^{-2}(-1)(-1) = -1 + \frac{1}{(1-x)^2}$
At the turning points, $\frac{dy}{dx} = 0$:
$-1 + \frac{1}{(1-x)^2} = 0 \implies (1-x)^2 = 1$
$1 - x = 1 \quad \text{or} \quad 1 - x = -1$
$x = 0 \quad \text{or} \quad x = 2$
Substitute back to find the corresponding $y$-values:
When $x = 0$, $y = -0 + 2 + \frac{1}{1-0} = 3$.
When $x = 2$, $y = -2 + 2 + \frac{1}{1-2} = -1$.

The turning points are $(0, 3)$ and $(2, -1)$.

4. Describe the sketch of the curve $C$.

Draw a set of axes and include the following:

• A vertical dashed line at $x = 1$.
• A dashed line for the oblique asymptote $y = -x + 2$.
• The upper branch for $x < 1$, passing through the local minimum and $y$-intercept at $(0, 3)$, bounded by the asymptotes.
• The lower branch for $x > 1$, with a local maximum at $(2, -1)$, bounded by the asymptotes.

1. Form a quadratic equation in $x$ for the intersection

At intersection points $kx = \dfrac{x^{2}-3x+3}{1-x}$.

Multiply by $(1-x)$:
$kx(1-x) = x^{2}-3x+3$
$kx - kx^{2} = x^{2}-3x+3$
Rearranging:
$(1+k)x^{2} - (3+k)x + 3 = 0.$
For two points of intersection this quadratic must have two distinct real roots, so we also need $1+k \neq 0$.

2. Apply discriminant > 0 and solve the quadratic inequality in $k$

Two distinct real roots require discriminant $> 0$:
$(3+k)^{2} - 4(1+k)(3) > 0$
$9 + 6k + k^{2} - 12 - 12k > 0$
$k^{2} - 6k - 3 > 0.$
Roots of $k^{2}-6k-3=0$: $k = \dfrac{6 \pm \sqrt{36+12}}{2} = 3 \pm 2\sqrt{3}$.

Since the parabola opens upward:
$k < 3 - 2\sqrt{3} \quad \text{or} \quad k > 3 + 2\sqrt{3}.$

3. Exclude the degenerate value and state the set

When $1+k = 0$, i.e. $k = -1$, the equation becomes linear $-2x+3 = 0$, giving only ONE intersection. Since $-1 < 3-2\sqrt{3}\ (\approx -0.46)$, $k=-1$ lies in the region above and must be removed.

Hence
$\{\,k \in \mathbb{R}: k < 3 - 2\sqrt{3} \text{ or } k > 3 + 2\sqrt{3}\,\} \setminus \{-1\}.$