Consider the points $A(\sqrt{a x},-\sqrt{b y})$, $B\left(\frac{n}{\sqrt{b}}, \frac{m}{\sqrt{a}}\right)$, and $\angle A O B=\theta$, then
$$\begin{array}{l} f(x, y)=O A \cdot O B \sin \theta \\ =\sqrt{(0-\sqrt{a x})^{2}+(0+\sqrt{b y})^{2}} \\ \cdot \sqrt{\left(0-\frac{n}{\sqrt{b}}\right)^{2}+\left(0-\frac{m}{\sqrt{a}}\right)^{2}} \cdot \sin \theta \\ =\sqrt{\frac{a n^{2}+b m^{2}}{a b}} \sqrt{c} \sin \theta . \end{array}$$

Let $k=\min \left(\frac{n}{\sqrt{b}},-\frac{m}{\sqrt{a}}\right)$, then
$$\begin{array}{l} k / \sqrt{\frac{a n^{2}+b m^{2}}{a b}} \leqslant \sin \theta \leqslant 1 . \\ \therefore f_{\text{max}}(x, y)=\sqrt{\frac{c}{a b}\left(a n^{2}+b m^{2}\right)}, \\ \quad f_{\text{min}}(x, y)=\sqrt{c} k . \end{array}$$