The correct option is:
$\text{For }\alpha =3,\underset{x\to x_0}{lim}\left|\frac{g\left(x\right)+e^{x_0}}{x-x_0}\right|=0$

Step-by-step Derivation:

Step 1: Understand the given condition
We are given that $x_0$ is a real number satisfying:
$e^{x_0}+x_0=0$
This implies:
$e^{x_0}=-x_0$

Step 2: Simplify the function $g\left(x\right)$
The function $g\left(x\right)$ is defined as:
$g\left(x\right)=\frac{3x{e}^x+3x-\alpha e^x-\alpha x}{3(e^x+1)}$
We can factor out terms in the numerator:
$g\left(x\right)=\frac{3x(e^x+1)-\alpha \left(e^{x}x\right)}{3(e^x+1)}$
Splitting the fraction gives:
$g\left(x\right)=x-\frac{\alpha }{3}\cdot \frac{e^x+x}{e^x+1}$

Step 3: Evaluate $g\left(x_0\right)$
Substitute $x=x_0$ into the expression for $g\left(x\right)$:
$g\left(x_0\right)=x_0-\frac{\alpha }{3}\cdot \frac{e^{x_0}+x_0}{e^{x_0}+1}$
Since $e^{x_0}+x_0=0$, we have:
$g\left(x_0\right)=x_0-0=x_0$
Using $x_0=-e^{x_0}$, we get:
$g\left(x_0\right)=-e^{x_0}\Rightarrow g\left(x_0\right)+e^{x_0}=0$

Step 4: Relate the limit to the derivative $g^{\prime }\left(x_0\right)$
Consider the limit:
$\underset{x\to x_0}{lim}\frac{g\left(x\right)+e^{x_0}}{x-x_0}$
Since $g\left(x_0\right)=-e^{x_0}$, this expression is:
$\underset{x\to x_0}{lim}\frac{g\left(x\right)-g\left(x_0\right)}{x-x_0}=g^{\prime }\left(x_0\right)$

Step 5: Differentiate $g\left(x\right)$
Let $h\left(x\right)=\frac{e^x+x}{e^x+1}$. Thus, $g\left(x\right)=x-\frac{\alpha }{3}h\left(x\right)$.
Using the quotient rule to differentiate $h\left(x\right)$:
$h^{\prime }\left(x\right)=\frac{(e^x+1)(e^x+1)-(e^x+x)e^x}{{(e^x+1)}^2}$
Evaluating $h^{\prime }\left(x_0\right)$:
Since $e^{x_0}+x_0=0$, the second term in the numerator becomes zero:
$h^{\prime }\left(x_0\right)=\frac{{(e^{x_0}+1)}^2-0}{{(e^{x_0}+1)}^2}=1$

Step 6: Compute $g^{\prime }\left(x_0\right)$ and find $\alpha$
Now, differentiate $g\left(x\right)$:
$g^{\prime }\left(x\right)=1-\frac{\alpha }{3}{h}^{\prime }\left(x\right)$
At $x=x_0$:
$g^{\prime }\left(x_0\right)=1-\frac{\alpha }{3}\left(1\right)=1-\frac{\alpha }{3}$
To make the limit equal to $0$:
$\left|g^{\prime }\right(x_0\left)\right|=0\Rightarrow 1-\frac{\alpha }{3}=0\Rightarrow \alpha =3$