Answer: \(y=3\left(\left(\frac{3}{2-x}\right)^2-1\right)\).

We are given

\(x=2-3\sin\theta.\)

Rearrange this to express \(\sin\theta\) in terms of \(x\):

\(3\sin\theta=2-x,\)

so

\(\sin\theta=\dfrac{2-x}{3}.\)

Also,

\(y=3\cot^2\theta.\)

Using

\(\cot^2\theta=\dfrac{\cos^2\theta}{\sin^2\theta}=\dfrac{1-\sin^2\theta}{\sin^2\theta},\)

we get

\(y=3\left(\dfrac{1-\sin^2\theta}{\sin^2\theta}\right).\)

Substitute \(\sin\theta=\dfrac{2-x}{3}\):

\(y=3\left(\dfrac{1-\left(\dfrac{2-x}{3}\right)^2}{\left(\dfrac{2-x}{3}\right)^2}\right).\)

This is the same as

\(y=3\left(\dfrac{1}{\left(\dfrac{2-x}{3}\right)^2}-1\right).\)

Therefore

\(y=3\left(\left(\dfrac{3}{2-x}\right)^2-1\right).\)