曲线方程由下式给出：
$Ax^2 + By^2 + Cx + Dy + 13 = 0$
其中 $A$、$B$、$C$ 和 $D$ 为常数。

1. Differentiate the given equation implicitly with respect to $x$.

$\frac{\mathrm{d}}{\mathrm{d}x}(Ax^2 + By^2 + Cx + Dy + 13) = \frac{\mathrm{d}}{\mathrm{d}x}(0)$
$2Ax + 2By\frac{\mathrm{d}y}{\mathrm{d}x} + C + D\frac{\mathrm{d}y}{\mathrm{d}x} = 0$

2. Isolate the terms with $\frac{\mathrm{d}y}{\mathrm{d}x}$ and factorize.

$2By\frac{\mathrm{d}y}{\mathrm{d}x} + D\frac{\mathrm{d}y}{\mathrm{d}x} = -2Ax - C$
$(2By + D)\frac{\mathrm{d}y}{\mathrm{d}x} = -(2Ax + C)$

3. Make $\frac{\mathrm{d}y}{\mathrm{d}x}$ the subject of the formula.

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{-(2Ax + C)}{2By + D}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{2Ax + C}{2By + D}$

1. Find the implicit derivative (from part i) and apply the stationary point condition to find a relationship between $A$ and $C$.

First, we find $\frac{\mathrm{d}y}{\mathrm{d}x}$ as required from the earlier part. Differentiating $Ax^2 + By^2 + Cx + Dy + 13 = 0$ implicitly with respect to $x$ gives:
$2Ax + 2By \frac{\mathrm{d}y}{\mathrm{d}x} + C + D \frac{\mathrm{d}y}{\mathrm{d}x} = 0$
$\frac{\mathrm{d}y}{\mathrm{d}x}(2By + D) = -(2Ax + C)$
$\frac{\mathrm{d}y}{\mathrm{d}x} = -\frac{2Ax + C}{2By + D}$
At the stationary point $(1, -1)$, $\frac{\mathrm{d}y}{\mathrm{d}x} = 0$. Setting the numerator to $0$:
$2A(1) + C = 0 \implies C = -2A$

2. Apply the condition for a tangent parallel to the $y$-axis to find a relationship between $B$ and $D$.

At $(3, -2)$, the tangent is parallel to the $y$-axis, meaning the gradient $\frac{\mathrm{d}y}{\mathrm{d}x}$ is undefined. Setting the denominator to $0$:
$2B(-2) + D = 0 \implies -4B + D = 0 \implies D = 4B$

3. Substitute both points into the curve's equation to form a system of equations in $A$ and $B$.

Since $(1, -1)$ lies on the curve:
$A(1)^2 + B(-1)^2 + C(1) + D(-1) + 13 = 0$
$A + B + C - D + 13 = 0$
Substitute $C = -2A$ and $D = 4B$:
$A + B - 2A - 4B + 13 = 0$
$-A - 3B + 13 = 0 \implies A = 13 - 3B \quad \text{--- (1)}$

Since $(3, -2)$ lies on the curve:
$A(3)^2 + B(-2)^2 + C(3) + D(-2) + 13 = 0$
$9A + 4B + 3C - 2D + 13 = 0$
Substitute $C = -2A$ and $D = 4B$:
$9A + 4B + 3(-2A) - 2(4B) + 13 = 0$
$9A + 4B - 6A - 8B + 13 = 0$
$3A - 4B + 13 = 0 \quad \text{--- (2)}$

4. Solve the system of linear equations to find the values of $A, B, C$, and $D$.

Substitute (1) into (2):
$3(13 - 3B) - 4B + 13 = 0$
$39 - 9B - 4B + 13 = 0$
$52 - 13B = 0 \implies B = 4$

Substitute $B = 4$ back into (1):
$A = 13 - 3(4) = 1$

Using our earlier expressions to find $C$ and $D$:
$C = -2(1) = -2$
$D = 4(4) = 16$