We are given two equations for $x$ and $y$ in terms of a parameter $t$:

$x = \csc^{-1}\left(\frac{1 + t^2}{2t}\right)$ and $y = \tan^{-1}\left(\frac{3t - t^3}{1 - 3t^2}\right).$

We need to find $\frac{dy}{dx}$. This involves using the chain rule, i.e.,

$\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx}.$

Step 1: Differentiate $x$ with respect to $t$

We are given: $x = \csc^{-1}\left(\frac{1 + t^2}{2t}\right).$

The derivative of $\csc^{-1}(u)$ with respect to $t$ is:

$\frac{d}{dt} \csc^{-1}(u) = -\frac{1}{|u|\sqrt{u^2 - 1}} \cdot \frac{du}{dt}.$

Here, $u = \frac{1 + t^2}{2t}$, so we first need to find $\frac{du}{dt}$:

$u = \frac{1 + t^2}{2t}.$ Using the quotient rule:

$\frac{du}{dt} = \frac{(2t)(2t) - (1 + t^2)(2)}{(2t)^2} = \frac{4t^2 - 2 - 2t^2}{4t^2} = \frac{2t^2 - 2}{4t^2} = \frac{1 - 1/t^2}{2}.$

Thus,

[ \frac{dx}{dt} = -\frac{1}{\