Before we start, let us recall a formula for the area of a triangle in terms of the lines bounding it.
Let $\hat{z}$ be the unit vector $(0,0,1)$.
For any line $\ell : \alpha x + \beta y + \gamma = 0$ on the plane, we will use
the same symbol $\ell$ to denote the affine function $\ell(x,y) = \alpha x + \beta y + \gamma$ and $\vec{\ell}$ to denote the vector $(\alpha,\beta,\gamma)$.
Given any three lines $\ell_i : \alpha_i x + \beta_i y + \gamma_i = 0$ for $i = 1,2,3$, the area of the triangle bounded by them is given by the formula: $$\verb/Area/ = \frac{D^2}{2C_1C_2C_3}$$
where
$$
D = \vec{\ell}_1 \cdot (\vec{\ell}_2 \times \vec{\ell}_3) = \left|\begin{matrix}
\alpha_1 & \alpha_2 & \alpha_3 \\
\beta_1 & \beta_2 & \beta_3 \\
\gamma_1 & \gamma_2 & \gamma_3
\end{matrix}\right|
\quad\text{ and }\quad
\begin{cases}
C_1 &= |\hat{z} \cdot (\vec{\ell}_2 \times \vec{\ell}_3 )| =
\left\|\begin{matrix}\alpha_2 & \alpha_3\\ \beta_2 & \beta_3\end{matrix}\right\|\\
C_2 &= |\hat{z} \cdot (\vec{\ell}_3 \times \vec{\ell}_1 )| =
\left\|\begin{matrix}\alpha_3 & \alpha_1\\ \beta_3 & \beta_1\end{matrix}\right\|\\
C_3 &= |\hat{z} \cdot (\vec{\ell}_1 \times \vec{\ell}_2 )| =
\left\|\begin{matrix}\alpha_1 & \alpha_2\\ \beta_1 & \beta_2\end{matrix}\right\|
\end{cases}
$$
For a derivation of this result, please refer to my
answer to a related
question.
Let's back to our original problem. For simplicity, we will only consider the case where all $\beta_i = 1$. The union of the three lines is the zero set of a cubic polynomial:
$$\begin{align}f(x,y) &= \ell_1(x,y)\ell_2(x,y)\ell_3(x,y)\\
&= y^3 + by^2x + cyx^2 + dx^3 + ey^2 + fyx + gx^2 + hy + ix + j
\end{align}
$$
Partial differentiate against $y$, we obtain
$$\begin{align}\partial_y f(x,y) &= \ell_1(x,y) \ell_2(x,y) + \ell_2(x,y)
\ell_3(x,y) + \ell_3(x,y) \ell_1(x,y)&\tag{*1a}\\
&= 3y^2 + 2bxy + 2ey + cx^2 + fx + h\\
&=
\begin{bmatrix}x \\ y \\ 1\end{bmatrix}^T
\begin{bmatrix}c & b & \frac{f}{2} \\ b & 3 & e\\ \frac{f}{2} & e & h\end{bmatrix}
\begin{bmatrix}x \\ y \\ 1\end{bmatrix}
& \tag{*1b}
\end{align}$$
Let $\Delta$ be the $3 \times 3$ matrix having $\vec{\ell}_i$ at $i^{th}$ column
and $\Lambda$ be the $3 \times 3$ matrix appear in $(*1b)$.
If we rewrite the RHS in $(*1a)$ in matrix form, we obtain
$$2\Lambda =
\left(\vec{\ell}_1 \vec{\ell}_2^T + \vec{\ell}_2 \vec{\ell}_1^T\right) +
\left(\vec{\ell}_2 \vec{\ell}_3^T + \vec{\ell}_3 \vec{\ell}_2^T\right) +
\left(\vec{\ell}_3 \vec{\ell}_1^T + \vec{\ell}_1 \vec{\ell}_3^T\right)
= \Delta \begin{bmatrix}0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0\end{bmatrix} \Delta^T
$$
This leads to
$$D^2 = (\det\Delta)^2 = \frac12\det(2\Lambda) = 12ch-4b^2h-3f^2+4bef-4ce^2\tag{*2}$$
Since all $\beta_i = 1$, we have
$$(C_1C_2C_3)^2 = (\alpha_2 - \alpha_3)^2(\alpha_3 - \alpha_1)^2(\alpha_1-\alpha_2)^2$$
Notice $-\alpha_i$ are roots of the cubic polynomial $\lambda^3 + b\lambda^2 + c\lambda + d = 0$. The expression on RHS is nothing but the discriminant of this cubic polynomial. i.e.
$$(C_1C_2C_3)^2 = b^2c^2 - 4c^3 - 4b^3d - 27d^2 + 18bcd\tag{*3}$$
Combine $(*2)$ and $(*3)$, we obtain
$$\verb/Area/ = \frac{12ch-4b^2h-3f^2+4bef-4ce^2}{2\sqrt{b^2c^2 - 4c^3 - 4b^3d - 27d^2 + 18bcd}}$$
As an example, consider the case
$$\begin{align} & x^2 y - y^3 - x^2 + 5y^2 - 8y + 4 = 0\\
\iff & y^3 - x^2 y - 5y^2 + x^2 + 8y - 4 = 0\\
\iff & (c,e,g,h,j) = (-1,-5,1,8,-4), b = d = f = i = 0
\end{align}
$$
Since $b = d = f = i = 0$, the area reduces to
$$\verb/Area/ \leadsto \frac{12ch-4ce^2}{2\sqrt{- 4c^3}} = \frac{12(-1)(8)-4(-1)(-5)^2}{2\sqrt{-4(-1)^3}} = \frac{4}{2\sqrt{4}} = 1$$
as expected.
