Step 1: Rewrite $g(x)$ in terms matching $f(x)$.
$g(x) = \int \dfrac{2\,dx}{e^{3x}+8e^x+4e^{-x}}$. Multiply numerator and denominator by $e^x$: \[ g(x) = \int \frac{2e^x\,dx}{e^{4x}+8e^{2x}+4}. \]
Step 2: Write $f(x)$ in the same denominator form.
$f(x) = \int \dfrac{e^{3x}}{4+8e^{2x}+e^{4x}}\,dx = \int \dfrac{e^{3x}}{e^{4x}+8e^{2x}+4}\,dx$.
Step 3: Compute $f(x) - g(x)$.
\[ f(x) - g(x) = \int \frac{e^{3x} - 2e^x}{e^{4x}+8e^{2x}+4}\,dx = \int \frac{e^x(e^{2x}-2)}{e^{4x}+8e^{2x}+4}\,dx. \]
Step 4: Substitute $u = e^x$.
$du = e^x\,dx$, so: \[ \int \frac{u^2 - 2}{u^4 + 8u^2 + 4}\,du. \] Divide numerator and denominator by $u^2$: \[ \int \frac{1 - 2/u^2}{u^2 + 8 + 4/u^2}\,du = \int \frac{1 - 2/u^2}{(u + 2/u)^2 + 4}\,du. \]
Step 5: Let $v = u + 2/u = e^x + 2e^{-x}$.
Then $dv = (1 - 2/u^2)\,du$. The integral becomes: \[ \int \frac{dv}{v^2 + 4} = \frac{1}{2}\tan^{-1}\left(\frac{v}{2}\right) + C = \frac{1}{2}\tan^{-1}\left(\frac{e^x + 2e^{-x}}{2}\right) + C. \]
Step 6: State the answer.
$f(x) - g(x) = \dfrac{1}{2}\tan^{-1}\!\left(\dfrac{e^x + 2e^{-x}}{2}\right) + C$.
\[ \boxed{\dfrac{1}{2}\tan^{-1}\!\left(\dfrac{e^x + 2e^{-x}}{2}\right) + C} \]