A curve has the equation $y = \frac{e^{2x+1}}{x+3}$, where $x \ne -3$.

1. Apply the quotient rule to differentiate the function

Let $u = e^{2x+1}$ and $v = x+3$.
Using the quotient rule $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{v \frac{\mathrm{d}u}{\mathrm{d}x} - u \frac{\mathrm{d}v}{\mathrm{d}x}}{v^2}$:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(x+3) \frac{\mathrm{d}}{\mathrm{d}x}(e^{2x+1}) - e^{2x+1} \frac{\mathrm{d}}{\mathrm{d}x}(x+3)}{(x+3)^2}$

2. Differentiate the numerator and denominator terms

$\frac{\mathrm{d}}{\mathrm{d}x}(e^{2x+1}) = 2e^{2x+1}$
$\frac{\mathrm{d}}{\mathrm{d}x}(x+3) = 1$
Substitute these into the quotient rule expression:
$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{(x+3)(2e^{2x+1}) - e^{2x+1}(1)}{(x+3)^2}$

3. Factorise and simplify the numerator

$\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{2x+1}[2(x+3) - 1]}{(x+3)^2}$
$= \frac{e^{2x+1}(2x + 6 - 1)}{(x+3)^2}$
$= \frac{e^{2x+1}(2x+5)}{(x+3)^2}$

1. State the derivative found in part (i)

From (i), $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{2x+1}(2x+5)}{(x+3)^2}$.

2. Analyse the sign of the derivative for $x > 0$

For $x > 0$:
$e^{2x+1} > 0$ (since exponential functions are always positive)
$2x+5 > 0$
$(x+3)^2 > 0$
Therefore, $\frac{\mathrm{d}y}{\mathrm{d}x} = \frac{e^{2x+1}(2x+5)}{(x+3)^2} > 0$.

3. Conclude whether the graph is increasing or decreasing

Since $\frac{\mathrm{d}y}{\mathrm{d}x} > 0$ for $x > 0$, the graph of $y = \frac{e^{2x+1}}{x+3}$ is increasing.