Let $f$ and $g$ be two functions defined by $f(x) = |x + |x||$ and $g(x) = \frac{1}{x}$ for $x \neq 0$. If $f(a) + g(f(a)) = \frac{13}{6}$ for some real $a$, then the maximum possible value of$f(g(a))$ is:

First, let's simplify $f(x) = |x + |x||$ by splitting into cases:

When $x \geq 0$: $|x| = x$, so $f(x) = |x + x| = 2x$

When $x < 0$: $|x| = -x$, so $f(x) = |x - x| = 0$

---

If $a < 0$, then $f(a) = 0$, and $g(f(a)) = g(0) = \dfrac{1}{0}$ which is undefined.

So $a$ must be positive, i.e., $a > 0$.

---

Since $a > 0$, we have $f(a) = 2a$. Substituting into the given equation:

$f(a) + g(f(a)) = \dfrac{13}{6}$

$2a + \dfrac{1}{2a} = \dfrac{13}{6}$

Let $K = 2a$ where $K > 0$:

$K + \dfrac{1}{K} = \dfrac{13}{6}$

$6K^2 - 13K + 6 = 0$

$(2K - 3)(3K - 2) = 0$

$K = \dfrac{3}{2}$ or $K = \dfrac{2}{3}$

Since $K = 2a$:

$a = \dfrac{3}{4}$ or $a = \dfrac{1}{3}$

---

Now, since $a > 0$, we get $g(a) = \dfrac{1}{a} > 0$, so:

$f(g(a)) = f\left(\dfrac{1}{a}\right) = \dfrac{2}{a}$

When $a = \dfrac{1}{3}$:

$f(g(a)) = \dfrac{2}{1/3} = 6$

When $a = \dfrac{3}{4}$:

$f(g(a)) = \dfrac{2}{3/4} = \dfrac{8}{3}$

Smaller $a$ gives a larger value of $\dfrac{2}{a}$, so the maximum occurs at $a = \dfrac{1}{3}$.

---

$\boxed{[f(g(a))]_{\max} = 6}$