Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function such that $f\left(\dfrac{x+y}{3}\right) = \dfrac{f(x) + f(y)}{3}$ for all $x, y \in \mathbb{R}$, and $f'(0) = 3$. Then the minimum value of the function $g(x) = 3 + e^x f(x)$, is:

Given $f\left(\dfrac{x+y}{3}\right) = \dfrac{f(x) + f(y)}{3}$

Substituting $x = 0$ and $y = 0$, we get:

$f(0) = \dfrac{2f(0)}{3} \Rightarrow f(0) = 0$

Differentiating the given equation partially with respect to $x$, treating $y$ as a constant:

$f'\left(\dfrac{x+y}{3}\right) \cdot \dfrac{1}{3} = \dfrac{f'(x)}{3}$

$\Rightarrow f'\left(\dfrac{x+y}{3}\right) = f'(x)$

Substituting $x = 0$, we get:

$f'\left(\dfrac{y}{3}\right) = f'(0) = 3$

Since this is true for all $y \in \mathbb{R}$, $f'(x) = 3$ for all $x \in \mathbb{R}$.

Integrating both sides with respect to $x$:

$f(x) = 3x + C$

Using $f(0) = 0$, we get $C = 0$. Thus, $f(x) = 3x$.

Now, the function $g(x)$ is given by:

$g(x) = 3 + e^x f(x) = 3 + 3x e^x$

To find the minimum value, we differentiate $g(x)$ with respect to $x$:

$g'(x) = 3(e^x + x e^x) = 3e^x(1 + x)$

Setting $g'(x) = 0$ gives $x = -1$.

For $x < -1$, $g'(x) < 0$ and for $x > -1$, $g'(x) > 0$. Therefore, $x = -1$ is a point of global minimum.

The minimum value of $g(x)$ is:

$g(-1) = 3 + 3(-1)e^{-1} = 3 - \dfrac{3}{e} = 3\left(\dfrac{e-1}{e}\right)$

Answer: $3\left(\dfrac{e-1}{e}\right)$