A curve $C$ has equation $y = \frac{a}{x+3} + bx + c$, where $a$, $b$ and $c$ are constants. It is given that $C$ passes through the point with coordinates $(1, -3)$ and has a stationary point $(-5, -21)$.

1. Formulate two equations by substituting the given coordinates into the curve's equation.

Since the curve passes through $(1, -3)$:
$-3 = \frac{a}{1+3} + b(1) + c$
$-3 = \frac{a}{4} + b + c$
$a + 4b + 4c = -12 \quad \text{--- (1)}$

Since the curve passes through $(-5, -21)$:
$-21 = \frac{a}{-5+3} + b(-5) + c$
$-21 = -\frac{a}{2} - 5b + c$
$-42 = -a - 10b + 2c$
$a + 10b - 2c = 42 \quad \text{--- (2)}$

2. Find the derivative $\frac{dy}{dx}$ and use the stationary point condition to form a third equation.

Given $y = a(x+3)^{-1} + bx + c$, differentiating with respect to $x$ gives:
$\frac{dy}{dx} = -a(x+3)^{-2} + b = -\frac{a}{(x+3)^2} + b$

At the stationary point $(-5, -21)$, $\frac{dy}{dx} = 0$:
$0 = -\frac{a}{(-5+3)^2} + b$
$0 = -\frac{a}{4} + b$
$a = 4b \quad \text{--- (3)}$

3. Solve the system of three linear equations simultaneously to find $a$, $b$, and $c$.

Substitute (3) into (1):
$(4b) + 4b + 4c = -12$
$8b + 4c = -12$
$2b + c = -3 \quad \text{--- (4)}$

Substitute (3) into (2):
$(4b) + 10b - 2c = 42$
$14b - 2c = 42$
$7b - c = 21 \quad \text{--- (5)}$

Add (4) and (5):
$(2b + c) + (7b - c) = -3 + 21$
$9b = 18 \implies b = 2$

Substitute $b = 2$ back into (4):
$2(2) + c = -3 \implies c = -7$

Substitute $b = 2$ into (3):
$a = 4(2) = 8$

1. Determine the values of $a$, $b$, and $c$ from the information given in part (i) to find the explicit equation of $C$.

From part (i), substitute $(1, -3)$ into $y$:
$\frac{a}{4} + b + c = -3$
Substitute $(-5, -21)$ into $y$:
$-\frac{a}{2} - 5b + c = -21$
Differentiate $y$ with respect to $x$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{a}{(x+3)^2} + b$
At the stationary point $(-5, -21)$, $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$:
$-\frac{a}{4} + b = 0 \implies a = 4b$
Solving the linear system yields $a = 8$, $b = 2$, and $c = -7$.
Thus, the equation of the curve is:
$y = \frac{8}{x+3} + 2x - 7$

2. Determine the equation of the vertical asymptote.

The term $\frac{8}{x+3}$ is undefined when $x + 3 = 0$.
As $x \to -3$, $y \to \pm\infty$.
Therefore, the vertical asymptote is $x = -3$.

3. Determine the equation of the oblique asymptote.

As $x \to \pm\infty$, the fractional term $\frac{8}{x+3} \to 0$.
Hence, the curve approaches $y = 2x - 7$.
Therefore, the oblique asymptote is $y = 2x - 7$.