1. We are given the function $$r=6(\sec \theta -\tan \theta)^{3/2}$$ and asked to find its derivative with respect to $$\theta$$.\n\n2. Let $$u=\sec \theta - \tan \theta$$, then $$r=6u^{3/2}$$. We'll use the chain rule: $$\frac{dr}{d\theta} = 6 \cdot \frac{3}{2} u^{1/2} \cdot \frac{du}{d\theta}$$.\n\n3. Next, find $$\frac{du}{d\theta}$$ where $$u=\sec \theta - \tan \theta$$. Using derivatives, $$\frac{d}{d\theta}(\sec \theta) = \sec \theta \tan \theta$$ and $$\frac{d}{d\theta}(\tan \theta) = \sec^{2} \theta$$, so $$\frac{du}{d\theta} = \sec \theta \tan \theta - \sec^{2} \theta$$.\n\n4. Substitute back into the derivative formula:\n$$\frac{dr}{d\theta} = 6 \cdot \frac{3}{2} (\sec \theta - \tan \theta)^{1/2} (\sec \theta \tan \theta - \sec^{2} \theta)$$\n\n5. Simplify the constants: $$6 \times \frac{3}{2} = 9$$. Thus,\n$$\boxed{\frac{dr}{d\theta} = 9 (\sec \theta - \tan \theta)^{1/2} (\sec \theta \tan \theta - \sec^{2} \theta)}$$\n\nThis is the derivative of the function in terms of $$\theta$$.