Let $f(x)=\frac{2-x^{2}}{6+x^{2}}$. Then,
\[
f^{\prime}(x)=
\]
\[
f^{\prime}(2)=
\]

Step 1 ：The derivative \(f'(x)\) is the rate of change of the function \(f(x)\) with respect to \(x\). We can find it using the quotient rule, which states that if \(f(x) = \frac{g(x)}{h(x)}\), then \(f'(x) = \frac{g'(x)h(x) - g(x)h'(x)}{[h(x)]^2}\).

Step 2 ：In this case, \(g(x) = 2 - x^2\) and \(h(x) = 6 + x^2\). So, \(g'(x) = -2x\) and \(h'(x) = 2x\).

Step 3 ：Substituting these into the quotient rule gives \(f'(x) = \frac{-2x(6 + x^2) - (2 - x^2)(2x)}{(6 + x^2)^2}\).

Step 4 ：Simplifying this expression gives \(f'(x) = \frac{-12x - 2x^3 - 4x + 2x^3}{(6 + x^2)^2} = \frac{-16x}{(6 + x^2)^2}\).

Step 5 ：Therefore, the derivative \(f'(x)\) is \(f'(x) = \frac{-16x}{(6 + x^2)^2}\).

Step 6 ：To find \(f'(2)\), we substitute \(x = 2\) into the derivative. This gives \(f'(2) = \frac{-16(2)}{(6 + 2^2)^2} = \frac{-32}{(6 + 4)^2} = \frac{-32}{100} = -0.32\).

Step 7 ：Therefore, \(f'(2) = \boxed{-0.32}\).