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

We start from the implicit relation

$$\left(a+\sqrt{2b}\cos x\right)\left(a-\sqrt{2b}\cos y\right)=a^{2}-b^{2},\qquad a>b>0.$$

Here both $$x$$ and $$y$$ are variables linked through the above equation, so we regard $$x$$ as a function of $$y$$ and differentiate with respect to $$y$$.

The right-hand side is a constant, so its derivative is $$0$$. For the left-hand side we first state the product rule:

$$\dfrac{d}{dy}[UV]=U\dfrac{dV}{dy}+V\dfrac{dU}{dy},$$

where $$U=a+\sqrt{2b}\cos x$$ and $$V=a-\sqrt{2b}\cos y$$. Applying the rule, we obtain

$$\bigl(a-\sqrt{2b}\cos y\bigr)\dfrac{d}{dy}\!\bigl(a+\sqrt{2b}\cos x\bigr) \;+\;\bigl(a+\sqrt{2b}\cos x\bigr)\dfrac{d}{dy}\!\bigl(a-\sqrt{2b}\cos y\bigr)=0.$$

Now we differentiate each factor. Remember that $$a$$ and $$b$$ are constants, while $$x$$ depends on $$y$$:

$$\dfrac{d}{dy}\!\bigl(a+\sqrt{2b}\cos x\bigr)=\sqrt{2b}\,(-\sin x)\dfrac{dx}{dy},$$

because $$\dfrac{d}{dy}\cos x=-\sin x\dfrac{dx}{dy}.$$

$$\dfrac{d}{dy}\!\bigl(a-\sqrt{2b}\cos y\bigr)=-\sqrt{2b}\,(-\sin y)=\sqrt{2b}\sin y,$$ because here $$y$$ is the differentiation variable itself.

Substituting these derivatives back gives

$$\bigl(a-\sqrt{2b}\cos y\bigr)\,\sqrt{2b}(-\sin x)\dfrac{dx}{dy} \;+\;\bigl(a+\sqrt{2b}\cos x\bigr)\,\sqrt{2b}\sin y=0.$$

Each term contains a common factor $$\sqrt{2b}$$; dividing by it simplifies the equation:

$$-\bigl(a-\sqrt{2b}\cos y\bigr)\sin x\,\dfrac{dx}{dy} \;+\;\bigl(a+\sqrt{2b}\cos x\bigr)\sin y=0.$$

Rearranging, we isolate $$\dfrac{dx}{dy}$$:

$$\dfrac{dx}{dy}=\dfrac{\bigl(a+\sqrt{2b}\cos x\bigr)\sin y} {\bigl(a-\sqrt{2b}\cos y\bigr)\sin x}.$$

We must now evaluate this derivative at the point $$\bigl(x,y\bigr)=\left(\dfrac{\pi}{4},\dfrac{\pi}{4}\right).$$ First record the basic trigonometric values

$$\cos\dfrac{\pi}{4}=\dfrac{\sqrt2}{2},\qquad \sin\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}.$$

Compute the auxiliary quantity

$$\sqrt{2b}\cos\dfrac{\pi}{4}=\sqrt{2b}\,\dfrac{\sqrt2}{2}=\sqrt{b}.$$

Hence at $$x=y=\dfrac{\pi}{4}$$ we have

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

and also

$$\sin x=\sin y=\dfrac{\sqrt2}{2}.$$

Substituting all these values into the derivative formula gives

$$\left.\dfrac{dx}{dy}\right|_{\left(\frac{\pi}{4},\frac{\pi}{4}\right)} =\dfrac{(a+\sqrt{b})\left(\dfrac{\sqrt2}{2}\right)} {(a-\sqrt{b})\left(\dfrac{\sqrt2}{2}\right)} =\dfrac{a+\sqrt{b}}{a-\sqrt{b}}.$$

At first glance the answer still contains $$\sqrt{b}$$, yet none of the options do. We therefore use the original relation to find an extra link between $$a$$ and $$b$$ at this specific point. Putting $$x=y=\dfrac{\pi}{4}$$ into the given equation itself, we get

$$\bigl(a+\sqrt{b}\bigr)\bigl(a-\sqrt{b}\bigr)=a^{2}-b^{2}.$$

The left side expands to $$a^{2}-b,$$ so

$$a^{2}-b=a^{2}-b^{2}\quad\Longrightarrow\quad b=b^{2} \;\Longrightarrow\; b(1-b)=0.$$

Because $$b>0$$, the only admissible root is $$b=1.$$ With this value, the derivative simplifies to

$$\dfrac{dx}{dy}=\dfrac{a+1}{a-1} =\dfrac{a+b}{a-b},$$

since now $$b=1.$$

This matches option C. Hence, the correct answer is Option C.