$\begin{array}{l}\text { 1. }\left[-1, \frac{5}{4}\right] . \\ \text { Let } x=2 \sec \theta, y=2 \tan \theta \text {. Then } \\ \frac{1}{x^{2}}+\frac{y}{x}=\frac{\cos ^{2} \theta}{4}-\sin \theta \\ =-\left(\frac{1}{2} \sin \theta+1\right)^{2}+\frac{5}{4} \\ \in\left[-1, \frac{5}{4}\right] .\end{array}$
1. If real numbers $x, y$ satisfy the condition $x^{2}-y^{2}=4$, then the range of $\frac{1}{x^{2}}-\frac{y}{x}$ is $\qquad$