Answer: \(y=\dfrac{1}{(3x-2)^2+1}\).

From

\(2+\cot\theta=3x\),

we have

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

Also, \(\sin\theta=\sqrt{y}\), so

\(\sin^2\theta=y\).

Use the identity

\(\csc^2\theta=1+\cot^2\theta\).

Since

\(\csc^2\theta=\dfrac{1}{\sin^2\theta}=\dfrac{1}{y}\),

we get

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

Therefore

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