1. Given equation for $x_0$

We are given $e^{x_0}+x_0=0 \,\Rightarrow\, e^{x_0}=-x_0.$

Since $e^{x_0}>0$, we must have $x_0<0$.

Also note that $e^{x_0}+1=1-x_0\neq 0,$ so $g(x)$ is well-defined at $x=x_0$.

2. Rewrite $g(x)$

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

Factor the numerator: $3xe^x+3x-\alpha e^x-\alpha x=(e^x+1)(3x-\alpha).$

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

So $g(x)$ is actually a very simple linear function.

3. Simplify the expression inside the limit

We need $\lim_{x\to x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|.$

Using $g(x)=x-\dfrac{\alpha}{3}$, $g(x)+e^{x_0}=x-\frac{\alpha}{3}+e^{x_0}.$

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

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

This form shows that unless the constant term in the numerator vanishes at $x=x_0$, the quotient will blow up. A cleaner way is to first compute $g(x_0)+e^{x_0}=\left(x_0-\frac{\alpha}{3}\right)+e^{x_0}=x_0-\frac{\alpha}{3}-x_0=-\frac{\alpha}{3}.$

Thus, for the limit to be finite, we need $g(x_0)+e^{x_0}=0 \iff -\frac{\alpha}{3}=0 \iff \alpha=0.$

But neither $\alpha=2$ nor $\alpha=3$ satisfies this. Therefore, for both given values, the quotient does not approach a finite real number; in fact its magnitude tends to $\infty$.

4. Check options

Option A: $\alpha=2$

Then $g(x)+e^{x_0}=x-\frac{2}{3}+e^{x_0}=(x-x_0)-\frac{2}{3},$ so near $x_0$, $\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|=\left|1-\frac{2}{3(x-x_0)}\right|\to \infty.$ Not $0$. So A is false.

Option B: $\alpha=2$

Same reasoning: limit is not $1$, it diverges. So B is false.

Option C: $\alpha=3$

Then $g(x)+e^{x_0}=x-1+e^{x_0}=(x-x_0)-1,$ so $\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|=\left|1-\frac{1}{x-x_0}\right|\to \infty.$ Not $0$. So C is false.

Option D: $\alpha=3$

Again the limit diverges, not $\frac23$. So D is false.

5. Conclusion

After simplification, $g(x)=x-\frac{\alpha}{3},$ and for both $\alpha=2$ and $\alpha=3$, $\lim_{x\to x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|$ does not equal any of the listed finite values; in fact it diverges to $\infty$.

So none of the options is true.

6. Comparison with stored answer

The stored correct answer is C, but our derivation shows C is false. Hence I do not agree with the stored answer.