It is given that $\mathrm{e}^y = 1 + \sin 3x$.

1. Differentiate the given equation $\mathrm{e}^y = 1 + \sin 3x$ implicitly with respect to $x$.

$\mathrm{e}^y \frac{\mathrm{d}y}{\mathrm{d}x} = 3 \cos 3x$

2. Differentiate the equation a second time using the product rule on the left side.

$\mathrm{e}^y \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \mathrm{e}^y \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = -9 \sin 3x$

3. Substitute $\sin 3x = \mathrm{e}^y - 1$ into the second derivative equation and rearrange to show the required identity.

$\mathrm{e}^y \left[ \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 \right] = -9(\mathrm{e}^y - 1)$
$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 = -9 + 9\mathrm{e}^{-y}$
$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + \left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)^2 + 9 = 9\mathrm{e}^{-y}$

4. Evaluate $y$, $\frac{\mathrm{d}y}{\mathrm{d}x}$, and $\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}$ at $x = 0$.

When $x=0$:
$\mathrm{e}^y = 1 + \sin 0 \implies \mathrm{e}^y = 1 \implies y = 0$
$\mathrm{e}^0 \frac{\mathrm{d}y}{\mathrm{d}x} = 3 \cos 0 \implies \frac{\mathrm{d}y}{\mathrm{d}x} = 3$
Substituting into the proven identity:
$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + (3)^2 + 9 = 9\mathrm{e}^0$
$\frac{\mathrm{d}^2 y}{\mathrm{d}x^2} + 18 = 9 \implies \frac{\mathrm{d}^2 y}{\mathrm{d}x^2} = -9$

5. Differentiate the proven identity with respect to $x$ to find an equation for $\frac{\mathrm{d}^3 y}{\mathrm{d}x^3}$ and evaluate it at $x=0$.

Differentiating both sides with respect to $x$:
$\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 2\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)\left(\frac{\mathrm{d}^2 y}{\mathrm{d}x^2}\right) = -9\mathrm{e}^{-y}\frac{\mathrm{d}y}{\mathrm{d}x}$
Substituting $x=0, y=0, y'=3, y''=-9$:
$\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} + 2(3)(-9) = -9(1)(3)$
$\frac{\mathrm{d}^3 y}{\mathrm{d}x^3} - 54 = -27 \implies \frac{\mathrm{d}^3 y}{\mathrm{d}x^3} = 27$

6. Construct the Maclaurin series expansion for $y$ up to the term in $x^3$.

$y = y(0) + y'(0)x + \frac{y''(0)}{2!}x^2 + \frac{y'''(0)}{3!}x^3 + \dots$
$y = 0 + 3x + \frac{-9}{2}x^2 + \frac{27}{6}x^3$
$y = 3x - \frac{9}{2}x^2 + \frac{9}{2}x^3$

1. Express $y$ explicitly and write the standard series for $\sin 3x$.

$\mathrm{e}^y = 1 + \sin 3x \implies y = \ln(1 + \sin 3x)$

Using the standard series for $\sin x$ from the List of Formulae:
$\sin 3x \approx 3x - \frac{(3x)^3}{3!} = 3x - \frac{27x^3}{6} = 3x - \frac{9}{2}x^3$

2. Substitute the expansion of $\sin 3x$ into the standard series for $\ln(1+u)$ up to the $x^3$ term.

Using the standard series for $\ln(1+x)$:
$\ln(1+u) \approx u - \frac{u^2}{2} + \frac{u^3}{3}$

Substitute $u = 3x - \frac{9}{2}x^3$:
$y = \left(3x - \frac{9}{2}x^3\right) - \frac{1}{2}\left(3x - \frac{9}{2}x^3\right)^2 + \frac{1}{3}\left(3x - \frac{9}{2}x^3\right)^3 + \dots$

3. Expand and simplify the terms up to $x^3$ to show it matches the expected result.

Expanding the powers, ignoring terms of degree $x^4$ and above:
$y \approx 3x - \frac{9}{2}x^3 - \frac{1}{2}(9x^2) + \frac{1}{3}(27x^3)$
$y \approx 3x - \frac{9}{2}x^2 + \left(-\frac{9}{2} + 9\right)x^3$
$y \approx 3x - \frac{9}{2}x^2 + \frac{9}{2}x^3$

This perfectly matches the series expansion obtained in part (a), verifying the result.