Given \(y = \frac{9}{2-x}\), we can rewrite it as \(y = 9(2-x)^{-1}\).

Using the chain rule, \(\frac{dy}{dx} = -9(2-x)^{-2} \times (-1)\).

This simplifies to \(\frac{dy}{dx} = \frac{9}{(2-x)^2}\).

For stationary points, \(\frac{dy}{dx} = 0\). However, \(\frac{9}{(2-x)^2} \neq 0\) for any real \(x\), so there are no stationary points.