if

$(a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^{2}-b^{2}$

Differentiating both sides,

$(-\sqrt{2} b \sin x)(a-\sqrt{2} b \cos y)+(a+\sqrt{2} b \cos x)$

$(\sqrt{2} b \sin y) \frac{d y}{d x}=0$

$\Rightarrow \frac{d y}{d x}=\frac{(\sqrt{2} b \sin x)(a-\sqrt{2} b \cos y)}{(a+\sqrt{2} b \cos x)(\sqrt{2} b \sin y)}$

$\therefore\left[\frac{d y}{d x}\right]\left(\frac{\pi}{4}, \frac{\pi}{4}\right)$

$=\frac{a-b}{a+b} \Rightarrow \frac{d x}{d y}=\frac{a+b}{a-b}$