The user wants me to solve a calculus problem by finding the first and second derivatives of a given function and evaluating them at specific points to check the given options.

Step 1: Find the first derivative, $f'(x)$.

The given function is $f(x) = \frac{x^2 - ax + 1}{x^2 + ax + 1}$. We can simplify the function by rewriting the numerator: $f(x) = \frac{x^2 + ax + 1 - 2ax}{x^2 + ax + 1} = 1 - \frac{2ax}{x^2 + ax + 1}$ Now, we differentiate $f(x)$ with respect to $x$ using the quotient rule for the fractional part. $f'(x) = -2a \frac{d}{dx}\left( \frac{x}{x^2 + ax + 1} \right)$ $f'(x) = -2a \left[ \frac{(1)(x^2 + ax + 1) - x(2x + a)}{(x^2 + ax + 1)^2} \right]$ $f'(x) = -2a \left[ \frac{x^2 + ax + 1 - 2x^2 - ax}{(x^2 + ax + 1)^2} \right]$ $f'(x) = -2a \left[ \frac{1 - x^2}{(x^2 + ax + 1)^2} \right]$ $f'(x) = \frac{2a(x^2 - 1)}{(x^2 + ax + 1)^2}$

Step 2: Evaluate $f'(1)$ and $f'(-1)$ and check options C and D.

Now, we evaluate the first derivative at $x=1$ and $x=-1$. At $x=1$: $f'(1) = \frac{2a(1^2 - 1)}{(1^2 + a(1) + 1)^2} = \frac{2a(0)}{(2+a)^2} = 0$ At $x=-1$: $f'(-1) = \frac{2a((-1)^2 - 1)}{((-1)^2 + a(-1) + 1)^2} = \frac{2a(0)}{(2-a)^2} = 0$ Now, let's check options C and D. Option C: $f'(1)f'(-1) = (2-a)^2$. We have $0 \cdot 0 = 0$. For the option to be true, $(2-a)^2$ must be 0, which implies $a=2$. However, the problem states $0 < a < 2$. So, option C is false. Option D: $f'(1)f'(-1) = -(2+a)^2$. We have $0 = -(2+a)^2$. This implies $a=-2$, which contradicts $0 < a < 2$. So, option D is false.

Step 3: Find the second derivative, $f''(x)$.

We differentiate $f'(x) = \frac{2a(x^2 - 1)}{(x^2 + ax + 1)^2}$ using the quotient rule. Let $u(x) = 2a(x^2-1)$ and $v(x) = (x^2+ax+1)^2$. Then $u'(x) = 4ax$ and $v'(x) = 2(x^2+ax+1)(2x+a)$. $f''(x) = \frac{u'(x)v(x) - u(x)v'(x)}{[v(x)]^2}$ $f''(x) = \frac{4ax(x^2+ax+1)^2 - 2a(x^2-1)[2(x^2+ax+1)(2x+a)]}{(x^2+ax+1)^4}$ We can cancel a factor of $(x^2+ax+1)$ from the numerator and denominator: $f''(x) = \frac{4ax(x^2+ax+1) - 4a(x^2-1)(2x+a)}{(x^2+ax+1)^3}$

Step 4: Evaluate $f''(1)$ and $f''(-1)$.

At $x=1$, the term $(x^2-1)$ is 0. So the second term in the numerator is 0. $f''(1) = \frac{4a(1)(1^2+a(1)+1) - 0}{(1^2+a(1)+1)^3} = \frac{4a(2+a)}{(2+a)^3} = \frac{4a}{(2+a)^2}$ At $x=-1$, the term $(x^2-1)$ is also 0. $f''(-1) = \frac{4a(-1)((-1)^2+a(-1)+1) - 0}{((-1)^2+a(-1)+1)^3} = \frac{-4a(1-a+1)}{(1-a+1)^3} = \frac{-4a(2-a)}{(2-a)^3} = \frac{-4a}{(2-a)^2}$

Step 5: Check options A and B.

Option A: ${\left( {2 + a} \right)^2}f''\left( 1 \right) + {\left( {2 - a} \right)^2}f''\left( { - 1} \right) = 0$ Let's substitute our results into the left-hand side (LHS): LHS = $(2+a)^2 \left( \frac{4a}{(2+a)^2} \right) + (2-a)^2 \left( \frac{-4a}{(2-a)^2} \right)$ LHS = $4a - 4a = 0$ Since LHS = 0, which is the right-hand side, option A is true.

Option B: ${\left( {2 - a} \right)^2}f''\left( 1 \right) - {\left( {2 + a} \right)^2}f''\left( { - 1} \right) = 0$ LHS = $(2-a)^2 \left( \frac{4a}{(2+a)^2} \right) - (2+a)^2 \left( \frac{-4a}{(2-a)^2} \right)$ LHS = $\frac{4a(2-a)^2}{(2+a)^2} + \frac{4a(2+a)^2}{(2-a)^2}$ Since $0 < a < 2$, both terms are positive, so their sum is positive and not equal to 0. Option B is false.

Final Answer is option A.