Differentiation

If $(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2$, where $\mathrm{a}>\mathrm{b}>0$, then $\frac{\mathrm{d} x}{\mathrm{~d} y}$ at $\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$ is