由 $17(x^{2}+y^{2})-30xy-16=0$，得$$(x+y)^{2}+16(x-y)^{2}=16,$$即$$\left(\dfrac{x+y}{4}\right)^{2}+(x-y)^{2}=1.$$令$$\begin{cases}x+y=4\cos\theta\\ x-y=\sin \theta\end{cases}(\theta\in\mathbb R)$$得\[\begin{split}f(x,y)&=\sqrt{16x^{2}+4y^{2}-16xy-12x+6y+9}\\&=\sqrt{(4x-2y)^{2}-3(4x-2y)+9}\\&=\sqrt{\left[3(x-y)+(x+y)\right]^{2}-3\left[3(x-y)+(x+y)\right]+9}\\&=\sqrt{\left(3\sin \theta+4\cos\theta\right)^{2}-3\left(3\sin\theta+4\cos\theta\right)+9}\\&=\sqrt{\left[5\sin(\theta+\alpha)\right]^{2}-3\left[5\sin(\theta+\alpha)\right]+9},\end{split}\]其中 $\alpha=\arcsin\dfrac{4}{5}$．
当 $\theta=\dfrac{3\pi}{2}-\alpha$ 时，$\sin(\theta+\alpha)=-1$，$f(x,y)$ 取得最大值 $7$．