Given that $y = \frac{\ln(x+3)^3}{3x+9}$ for $x > -3$, find the set of values of $x$ for which $y$ is an increasing function of $x$. Give your answer in terms of e.

1. Simplify the expression for $y$ using laws of logarithms.

$y = \frac{\ln(x+3)^3}{3x+9}$
$y = \frac{3\ln(x+3)}{3(x+3)}$
$y = \frac{\ln(x+3)}{x+3}$

2. Differentiate $y$ with respect to $x$ using the quotient rule.

Let $u = \ln(x+3)$ and $v = x+3$.
$\frac{du}{dx} = \frac{1}{x+3}$
$\frac{dv}{dx} = 1$
$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
$\frac{dy}{dx} = \frac{(x+3)\left(\frac{1}{x+3}\right) - \ln(x+3)(1)}{(x+3)^2}$
$\frac{dy}{dx} = \frac{1 - \ln(x+3)}{(x+3)^2}$

3. Set up the inequality for an increasing function.

For $y$ to be an increasing function of $x$, $\frac{dy}{dx} > 0$.
$\frac{1 - \ln(x+3)}{(x+3)^2} > 0$

4. Solve the inequality for $x$, taking into account the given domain.

Since $(x+3)^2 > 0$ for all $x > -3$, we must have:
$1 - \ln(x+3) > 0$
$\ln(x+3) < 1$
$x+3 < \text{e}^1$
$x < \text{e} - 3$

Given the domain $x > -3$, the set of values is:
$-3 < x < \text{e} - 3$