Let A(x) = pmatrix f(x) & g(x) -g(x) & f(x) pmatrix be a matrix...

Given $A(x+y) = A(x)A(y)$.
Differentiating partially with respect to $y$ and setting $y = 0$:
$A'(x) = A(x)A'(0)$

We are given $f'(0) = 2$ and $g'(0) = 3$, so:
$A'(0) = \begin{pmatrix} 2 & 3 \\ -3 & 2 \end{pmatrix}$

Thus, $\begin{pmatrix} f'(x) & g'(x) \\ -g'(x) & f'(x) \end{pmatrix} = \begin{pmatrix} f(x) & g(x) \\ -g(x) & f(x) \end{pmatrix} \begin{pmatrix} 2 & 3 \\ -3 & 2 \end{pmatrix}$

Equating the elements of the first row, we get:
$f'(x) = 2f(x) - 3g(x)$
$g'(x) = 3f(x) + 2g(x)$

We need to find $H(x) = (f'(x))^2 + (g'(x))^2$.
$H(x) = (2f(x) - 3g(x))^2 + (3f(x) + 2g(x))^2$
$H(x) = 4f(x)^2 - 12f(x)g(x) + 9g(x)^2 + 9f(x)^2 + 12f(x)g(x) + 4g(x)^2 = 13(f(x)^2 + g(x)^2)$

Let $V(x) = f(x)^2 + g(x)^2$. Differentiating with respect to $x$:
$V'(x) = 2f(x)f'(x) + 2g(x)g'(x)$
$V'(x) = 2f(x)(2f(x) - 3g(x)) + 2g(x)(3f(x) + 2g(x)) = 4f(x)^2 - 6f(x)g(x) + 6f(x)g(x) + 4g(x)^2 = 4(f(x)^2 + g(x)^2) = 4V(x)$

Solving the differential equation $V'(x) = 4V(x)$, we get:
$V(x) = V(0)e^{4x}$

From $A(0+0) = A(0)A(0)$, we have $A(0) = I$, which gives $f(0) = 1$ and $g(0) = 0$.
So, $V(0) = 1^2 + 0^2 = 1$.
Thus, $V(x) = e^{4x}$.

Therefore, $H(x) = 13e^{4x}$.
At $x_0 = \dfrac{1}{2}\ln(3)$:
$H(x_0) = 13e^{4 \times \dfrac{1}{2}\ln(3)} = 13e^{2\ln(3)} = 13e^{\ln(9)} = 13 \times 9 = 117$.

Answer: $117$