1. Given relation

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

We need to find $\dfrac{dx}{dy}$ at $\left(\dfrac{\pi}{4},\dfrac{\pi}{4}\right)$.

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2. Differentiate implicitly with respect to $y$

Treat $x$ as a function of $y$.

Let

$$U=a+\sqrt{2}\,b\cos x, \qquad V=a-\sqrt{2}\,b\cos y$$

Then

$$UV=a^2-b^2$$

Differentiating w.r.t. $y$:

$$U'V+UV'=0$$

Now,

$$U'=\sqrt{2}\,b(-\sin x)\frac{dx}{dy}=-\sqrt{2}\,b\sin x\frac{dx}{dy}$$

and

$$V'=-\sqrt{2}\,b(-\sin y)=\sqrt{2}\,b\sin y$$

So,

$$(-\sqrt{2}\,b\sin x\,\frac{dx}{dy})(a-\sqrt{2}\,b\cos y) +(a+\sqrt{2}\,b\cos x)(\sqrt{2}\,b\sin y)=0$$

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3. Solve for $\dfrac{dx}{dy}$

Divide throughout by $\sqrt{2}\,b$:

$$-\sin x\left(a-\sqrt{2}\,b\cos y\right)\frac{dx}{dy} +\left(a+\sqrt{2}\,b\cos x\right)\sin y=0$$

Hence,

$$\sin x\left(a-\sqrt{2}\,b\cos y\right)\frac{dx}{dy} =\left(a+\sqrt{2}\,b\cos x\right)\sin y$$

Therefore,

$$\frac{dx}{dy}= \frac{(a+\sqrt{2}\,b\cos x)\sin y}{\sin x\,(a-\sqrt{2}\,b\cos y)}$$

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4. Substitute $x=y=\dfrac{\pi}{4}$

We know:

$$\sin\frac{\pi}{4}=\frac{1}{\sqrt{2}}, \qquad \cos\frac{\pi}{4}=\frac{1}{\sqrt{2}}$$

So,

$$\sqrt{2}\,b\cos\frac{\pi}{4}=\sqrt{2}\,b\cdot \frac{1}{\sqrt{2}}=b$$

Thus,

$$\frac{dx}{dy}= \frac{(a+b)\cdot \frac{1}{\sqrt{2}}}{\frac{1}{\sqrt{2}}\cdot (a-b)} =\frac{a+b}{a-b}$$

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5. Match with options

$$\frac{dx}{dy}=\frac{a+b}{a-b}$$

This is Option C.

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6. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

So they agree.