The equation of a curve is $y = \frac{e^{2x}}{1-x}$, where $x > \frac{3}{2}$. Find $\frac{dy}{dx}$ and explain whether $y$ is always increasing or decreasing.

1. Differentiate $y$ using the quotient rule.

Let $u = e^{2x}$ and $v = 1-x$.
Then $\frac{du}{dx} = 2e^{2x}$ and $\frac{dv}{dx} = -1$.

$\frac{dy}{dx} = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
$\frac{dy}{dx} = \frac{(1-x)(2e^{2x}) - (e^{2x})(-1)}{(1-x)^2}$

2. Simplify the expression for $\frac{dy}{dx}$.

$\frac{dy}{dx} = \frac{2e^{2x} - 2xe^{2x} + e^{2x}}{(1-x)^2}$
$\frac{dy}{dx} = \frac{3e^{2x} - 2xe^{2x}}{(1-x)^2}$
$\frac{dy}{dx} = \frac{e^{2x}(3 - 2x)}{(1-x)^2}$

3. Analyse the sign of $\frac{dy}{dx}$ for $x > \frac{3}{2}$.

For $x > \frac{3}{2}$:
$2x > 3 \implies 3 - 2x < 0$
$e^{2x} > 0$ for all real $x$
$(1-x)^2 > 0$ for all $x \neq 1$

Therefore, the numerator $e^{2x}(3 - 2x) < 0$ and the denominator $(1-x)^2 > 0$.

4. Conclude whether $y$ is increasing or decreasing.

Since $\frac{dy}{dx} < 0$ for all $x > \frac{3}{2}$, $y$ is always decreasing.