A curve has parametric equations $x = 2t - \frac{1}{t}$ and $y = t + \frac{1}{t}$ where $t$ is a parameter and $t \ne 0$.

1. Find the derivatives of $x$ and $y$ with respect to the parameter $t$.

Given $x = 2t - t^{-1}$ and $y = t + t^{-1}$:
$\frac{dx}{dt} = 2 + t^{-2} = \frac{2t^2+1}{t^2}$
$\frac{dy}{dt} = 1 - t^{-2} = \frac{t^2-1}{t^2}$

2. Apply the chain rule for parametric equations to find $\frac{dy}{dx}$.

$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$
$\frac{dy}{dx} = \frac{\frac{t^2-1}{t^2}}{\frac{2t^2+1}{t^2}} = \frac{t^2-1}{2t^2+1}$

3. Manipulate the algebraic fraction to match the target form.

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

1. Establish the bounds for the fractional term involving $t$.

Since $t \neq 0$, $t^2 > 0$.
This implies that $2t^2 + 1 > 1$.
Taking the reciprocal gives:
$0 < \frac{1}{2t^2+1} < 1$
Multiplying by $\frac{3}{2}$:
$0 < \frac{3}{2(2t^2+1)} < \frac{3}{2}$

2. Determine the range of $\frac{dy}{dx}$ by constructing the full expression.

Multiplying the inequality by $-1$ reverses the inequality signs:
$-\frac{3}{2} < -\frac{3}{2(2t^2+1)} < 0$
Adding $\frac{1}{2}$ to all parts of the inequality:
$\frac{1}{2} - \frac{3}{2} < \frac{1}{2} - \frac{3}{2(2t^2+1)} < \frac{1}{2}$
$-1 < \frac{dy}{dx} < \frac{1}{2}$