7. Let $\bar{a}$ be a vector with coordinates $\bar{a}(3,-4,-\sqrt{11})$, and $\bar{b}$ be a vector with coordinates $\bar{b}(\operatorname{Sin} 2 x \operatorname{Cos} y, \operatorname{Cos} 2 x \operatorname{Cos} y, \operatorname{Siny})$. The length $|\bar{a}|$ is 6, and the length of vector $\bar{b}$ is clearly 1. Since $f(x, y)=\bar{a} \cdot \bar{b}$, where $\bar{a} \cdot \bar{b}$ is the dot product of the vectors, by the properties of the dot product, $-|\bar{a}||\bar{b}| \leq \bar{a} \cdot \bar{b} \leq|\bar{a}||\bar{b}|$. Therefore, $-6 \leq f(x, y) \leq 6$. Equality will be achieved if vectors $\bar{a}$ and $\bar{b}$ are collinear, i.e., $\frac{\operatorname{Sin} 2 x \operatorname{Cos} y}{3}=\frac{\operatorname{Cos} 2 x \operatorname{Cos} y}{-4}=\frac{\operatorname{Siny}}{-\sqrt{11}}$, which will be satisfied for corresponding $x$ and $y$. Thus, $\min f(x, y)=-6, \max f(x, y)=6$.
Answer: $\min f(x, y)=-6, \max f(x, y)=6$.