The curve $C$ has equation $y = \frac{x^2}{x - 1}$. Points $A(x_1, y_1)$ and $B(x_2, y_2)$ lie on curve $C$ such that the tangent at $A$ is parallel to tangent at $B$ where $x_2 < x_1$. Given further that the equation of tangent at $A$ is $y = \frac{8}{9}x + \frac{16}{9}$, find the coordinates of $B$, and hence find the equation of normal at point $B$.

1. Find the derivative of the curve equation.

$y = \frac{x^2}{x-1}$
Using the quotient rule:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(x-1)(2x) - x^2(1)}{(x-1)^2}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2x^2 - 2x - x^2}{(x-1)^2} = \frac{x^2 - 2x}{(x-1)^2}$

2. Equate the derivative to the gradient of the given tangent to solve for possible $x$-coordinates.

The equation of tangent at $A$ is $y = \frac{8}{9}x + \frac{16}{9}$, so its gradient is $\frac{8}{9}$.
Since the tangent at $B$ is parallel to the tangent at $A$, both have a gradient of $\frac{8}{9}$.
Equating $\frac{\mathrm{d}y}{\mathrm{d}x}$ to $\frac{8}{9}$:
$\frac{x^2 - 2x}{(x-1)^2} = \frac{8}{9}$
$9(x^2 - 2x) = 8(x^2 - 2x + 1)$
$9x^2 - 18x = 8x^2 - 16x + 8$
$x^2 - 2x - 8 = 0$

3. Solve for the coordinates of $B$.

$(x-4)(x+2) = 0$
$x = 4 \text{ or } x = -2$
Since $x_2 < x_1$, $x_2 = -2$ and $x_1 = 4$.
Substitute $x = -2$ into the curve equation to find $y_2$:
$y_2 = \frac{(-2)^2}{-2 - 1} = \frac{4}{-3} = -\frac{4}{3}$
Coordinates of $B$ are $\left(-2, -\frac{4}{3}\right)$.

4. Find the gradient of the normal at $B$.

Gradient of tangent at $B$, $m_T = \frac{8}{9}$.
Gradient of normal at $B$, $m_N = -\frac{1}{m_T} = -\frac{9}{8}$.

5. Form the equation of the normal at $B$.

Using the point-gradient form $y - y_1 = m(x - x_1)$:
$y - \left(-\frac{4}{3}\right) = -\frac{9}{8}(x - (-2))$
$y + \frac{4}{3} = -\frac{9}{8}x - \frac{9}{4}$
$y = -\frac{9}{8}x - \frac{9}{4} - \frac{4}{3}$
$y = -\frac{9}{8}x - \frac{43}{12}$