If ( a +√ 2 b cos x)( a -√ 2 b cos y)= a^2- b^2, where a > b >0, then (d x)/(d y) at ((π)/4, (π)/4) is:

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