We are given the equation \( f(x + ay) + g(x - ay) = 0 \). Differentiating both sides with respect to \( x \), we apply the chain rule:
\[
\frac{d}{dx}\left(f(x + ay)\right) + \frac{d}{dx}\left(g(x - ay)\right) = 0
\]
This gives:
\[
f'(x + ay) \cdot \frac{d}{dx}(x + ay) + g'(x - ay) \cdot \frac{d}{dx}(x - ay) = 0
\]
\[
f'(x + ay) \cdot (1 + a \frac{dy}{dx}) + g'(x - ay) \cdot (1 - a \frac{dy}{dx}) = 0
\]
Now, solve for \( \frac{dy}{dx} \):
\[
f'(x + ay) + a f'(x + ay) \frac{dy}{dx} + g'(x - ay) - a g'(x - ay) \frac{dy}{dx} = 0
\]
Rearrange the terms to isolate \( \frac{dy}{dx} \):
\[
a(f'(x + ay) + g'(x - ay)) \frac{dy}{dx} = -(f'(x + ay) + g'(x - ay))
\]
Thus:
\[
\frac{dy}{dx} = \frac{f'(x + ay) + g'(x - ay)}{g'(x - ay) - f'(x + ay)}
\]
Thus, the correct answer is option (2).