Step 1: Simplify $g(x)$.

Factor the numerator as $3x(e^x+1)-\alpha (e^x+x)$, giving $g(x)=x-\frac{\alpha }{3}\cdot \frac{e^x+x}{e^x+1}$.

Step 2: Use the defining relation.

Since $e^{x_0}+x_0=0$, we have $e^{x_0}=-x_0$, so $x+e^{x_0}=x-x_0$. Therefore

$$g(x)+e^{x_0}=(x-x_0)-\frac{\alpha }{3}\cdot \frac{e^x+x}{e^x+1}.$$

Step 3: Recognise the limit as $|g^{\prime }(x_0)|$.

Dividing by $(x-x_0)$ and taking the limit gives $\left|g^{\prime }(x_0)\right|$ (since the value of $g$ at $x_0$ is $-e^{x_0}$, the quotient is a difference-quotient form).

Step 4: Compute $g^{\prime }(x_0)$.

Differentiating, $g^{\prime }(x)=1-\frac{\alpha }{3}\cdot \frac{(e^x+1{)}^2-(e^x+x)e^x}{(e^x+1{)}^2}$. At $x=x_0$ the term $e^{x_0}+x_0$ vanishes, so the bracket collapses to $(e^{x_0}+1{)}^2$, leaving $g^{\prime }(x_0)=1-\frac{\alpha }{3}$.

Step 5: Evaluate for each $\alpha$.

For $\alpha =3$: $g^{\prime }(x_0)=0$, hence the limit is $0$. For $\alpha =2$: the limit equals $\frac{1}{3}$, not $0$ or $1$.

$\therefore$ The correct option is (C).