已知曲线 $y = \frac{1}{2}e^{2x} - \frac{1}{3}e^{-3x} + \frac{1}{3}$，且该曲线在点 $P\left(\ln 2, \frac{55}{24}\right)$ 处的法线与 $y$ 轴交于点 $Q$。

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

$y = \frac{1}{2}e^{2x} - \frac{1}{3}e^{-3x} + \frac{1}{3}$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{1}{2}(2e^{2x}) - \frac{1}{3}(-3e^{-3x}) + 0$

$\frac{\mathrm{d}y}{\mathrm{d}x} = e^{2x} + e^{-3x}$

2. Express the derivative as a single fraction to match the required form.

$\frac{\mathrm{d}y}{\mathrm{d}x} = e^{2x} + \frac{1}{e^{3x}}$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{2x} \cdot e^{3x} + 1}{e^{3x}}$

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{5x} + 1}{e^{3x}}$ (Shown)

1. Evaluate the gradient of the tangent at $x = \ln 2$.

From part (a), $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{5x} + 1}{e^{3x}}$.
When $x = \ln 2$,
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{5\ln 2} + 1}{e^{3\ln 2}}$
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{2^5 + 1}{2^3} = \frac{32 + 1}{8} = \frac{33}{8}$

2. Find the gradient of the normal.

Gradient of normal $m_n = -\frac{1}{\frac{33}{8}} = -\frac{8}{33}$

3. Find the equation of the normal.

Using $y - y_1 = m(x - x_1)$ with $P\left(\ln 2, \frac{55}{24}\right)$:
$y - \frac{55}{24} = -\frac{8}{33}(x - \ln 2)$
$y = -\frac{8}{33}x + \frac{8}{33}\ln 2 + \frac{55}{24}$

1. Find the gradient of the tangent at $P$ using the derivative from part (a)

At $x = \ln 2$, $e^x = 2$.
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{5(\ln 2)} + 1}{e^{3(\ln 2)}} = \frac{2^5 + 1}{2^3} = \frac{32 + 1}{8} = \frac{33}{8}$

2. Find the equation of the normal at $P$

Gradient of normal $m_n = -\frac{1}{\frac{33}{8}} = -\frac{8}{33}$
Equation of normal: $y - \frac{55}{24} = -\frac{8}{33}(x - \ln 2)$

3. Find the coordinates of $Q$ by substituting $x = 0$ into the equation of the normal

Since $Q$ lies on the $y$-axis, $x = 0$.
$y - \frac{55}{24} = -\frac{8}{33}(0 - \ln 2)$
$y = \frac{8}{33}\ln 2 + \frac{55}{24}$
Coordinates of $Q$ are $\left(0, \frac{8}{33}\ln 2 + \frac{55}{24}\right)$