Analyze Sign Variation of Derivative of Integral Function

To determine the sign variation of $g'(x)$, we first calculate the derivative $g'(x)$ using Leibniz's Rule for differentiation under the integral sign.

Given:
$g(x) = \int_{0}^{e^x} \frac{f'(t)}{1+t^2} dt$

Applying Leibniz's Rule with respect to $x$:
$g'(x) = \frac{f'(e^x)}{1 + (e^x)^2} \cdot \frac{d}{dx}\left(e^x\right) - 0$
$g'(x) = \frac{e^x f'(e^x)}{1 + e^{2x}}$

Since $e^x > 0$ and $1 + e^{2x} > 0$ for all $x \in (-\infty, \infty)$, the sign of $g'(x)$ is completely determined by the sign of $f'(e^x)$.

Next, we calculate $f'(t)$ using the quotient rule for $f(t) = \frac{t^2 - at + 1}{t^2 + at + 1}$:
$f'(t) = \frac{(2t - a)(t^2 + at + 1) - (t^2 - at + 1)(2t + a)}{(t^2 + at + 1)^2}$

Expanding the numerator:
$(2t - a)(t^2 + at + 1) = 2t^3 + at^2 + (2 - a^2)t - a$
$(t^2 - at + 1)(2t + a) = 2t^3 - at^2 + (2 - a^2)t + a$

Subtracting the two expressions gives:
$\text{Numerator} = 2at^2 - (-2at^2) - a - a = 2at^2 - 2a = 2a(t^2 - 1)$

Thus, the derivative $f'(t)$ simplifies to:
$f'(t) = \frac{2a(t^2 - 1)}{(t^2 + at + 1)^2}$

Now, substituting $t = e^x$:
$f'(e^x) = \frac{2a(e^{2x} - 1)}{(e^{2x} + a e^x + 1)^2}$

Since $0 < a < 2$, we have $2a > 0$ and $(e^{2x} + a e^x + 1)^2 > 0$ for all real numbers $x$. Hence, the sign of $f'(e^x)$, and consequently the sign of $g'(x)$, depends solely on the expression $(e^{2x} - 1)$:

1. For $x \in (-\infty, 0)$:
   $x < 0 \implies 2x < 0 \implies e^{2x} < 1 \implies e^{2x} - 1 < 0$
   Therefore, $g'(x) < 0$ on $(-\infty, 0)$ (i.e., $g'(x)$ is negative on $(-\infty, 0)$).

2. For $x \in (0, \infty)$:
   $x > 0 \implies 2x > 0 \implies e^{2x} > 1 \implies e^{2x} - 1 > 0$
   Therefore, $g'(x) > 0$ on $(0, \infty)$ (i.e., $g'(x)$ is positive on $(0, \infty)$).

Hence, $g'(x)$ is negative on $(-\infty, 0)$ and positive on $(0, \infty)$.

Correct Option: (B)