If \((a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2\), where \(\mathrm{a}\gt\mathrm{b}\gt0\), then \(\frac{\mathrm{d} x}{\mathrm{~d} y}\) at \(\left(\frac{\pi}{4}, \frac{\pi}{4}\right)\) is

\((a+\sqrt{2} b \cos x)(a-\sqrt{2} b \cos y)=a^2-b^2\)
Differentiating both sides w.r.t. \(y\), we get
\(\begin{aligned}
& \begin{aligned}
&(\mathrm{a}+\sqrt{2} \mathrm{~b} \cos x)(\sqrt{2} \mathrm{~b} \sin y) \\
& \quad+(\mathrm{a}-\sqrt{2} \mathrm{~b} \cos y)\left(-\sqrt{2} \mathrm{~b} \sin x \frac{\mathrm{~d} x}{\mathrm{~d} y}\right)=0 \\
& \Rightarrow \frac{\mathrm{~d} x}{\mathrm{~d} y}=\frac{\sqrt{2} \mathrm{~b} \sin y(\mathrm{a}+\sqrt{2} \mathrm{~b} \cos x)}{\sqrt{2} \mathrm{~b} \sin x(\mathrm{a}-\sqrt{2} \mathrm{~b} \cos y)} \\
& \Rightarrow\left(\frac{\mathrm{d} x}{\mathrm{~d} y}\right)_{\left(\frac{\pi}{4} \frac{\pi}{4}\right)}=\frac{\mathrm{b}(\mathrm{a}+\mathrm{b})}{\mathrm{b}(\mathrm{a}-\mathrm{b})}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{a}-\mathrm{b}}
\end{aligned}
\end{aligned}\)