Let $$x_0$$ be the real number such that $$e^{x_0} + x_0 = 0$$. For a given real number $$\alpha$$, define

$$g(x) = \frac{3xe^x + 3x - ae^x - ax}{3(e^x + 1)}$$

for all real numbers $$x$$.

Then which one of the following statements is TRUE?

We are given the unique real number $$x_0$$ that satisfies $$e^{x_0}+x_0=0$$ (hence $$x_0\lt 0$$ and $$e^{x_0}=-x_0$$).
For a real parameter $$\alpha$$, define

$$g(x)=\frac{3x\,e^{x}+3x-\alpha\,e^{x}-\alpha x}{3(e^{x}+1)},\qquad x\in\mathbb{R}.$$

Consider the function $$h(x)=g(x)+e^{x_0}.$$ The required limit is

$$L=\lim_{x\to x_0}\left|\frac{h(x)}{x-x_0}\right|.$$

If $$h(x_0)=0,$$ then by the first-principle definition of the derivative, the limit equals $$|h'(x_0)|=|g'(x_0)|.$$
We therefore first evaluate $$h(x_0).$$

Put $$e^{x_0}=-x_0$$ in $$g(x):$$

Numerator at $$x=x_0$$:

$$\begin{aligned} N_0&=3x_0e^{x_0}+3x_0-\alpha e^{x_0}-\alpha x_0\\ &=3x_0(-x_0)+3x_0-\alpha(-x_0)-\alpha x_0\\ &=3x_0(1-x_0). \end{aligned}$$

Denominator at $$x=x_0$$:

$$D_0=3(e^{x_0}+1)=3(-x_0+1)=3(1-x_0).$$

Thus $$g(x_0)=\dfrac{3x_0(1-x_0)}{3(1-x_0)}=x_0.$$

Hence $$h(x_0)=g(x_0)+e^{x_0}=x_0+(-x_0)=0.$$
Therefore $$L=|g'(x_0)|.$$

Next we differentiate $$g(x).$$ Write $$g(x)=\dfrac{U(x)}{V(x)}$$ with

$$U(x)=3x\,e^{x}+3x-\alpha e^{x}-\alpha x,\qquad V(x)=3(e^{x}+1)=3e^{x}+3.$$

Derivatives:

$$\begin{aligned} U'(x)&=3e^{x}(1+x)+3-\alpha e^{x}-\alpha\\ &=e^{x}\!\left[3(1+x)-\alpha\right]+(3-\alpha),\\ V'(x)&=3e^{x}. \end{aligned}$$

Using the quotient rule,

$$g'(x)=\frac{U'(x)V(x)-U(x)V'(x)}{[V(x)]^{2}}.$$

Evaluate each piece at $$x=x_0$$, remembering $$e^{x_0}=-x_0$$:

$$\begin{aligned} U'(x_0)&=(-x_0)\!\left[3(1+x_0)-\alpha\right]+(3-\alpha),\\ V(x_0)&=3(1-x_0),\\ V'(x_0)&=-3x_0,\\ U(x_0)&=3x_0(1-x_0). \end{aligned}$$

Substituting into the quotient-rule formula and simplifying (factor $$3(1-x_0)$$ and cancel with the squared denominator) gives

$$g'(x_0)=\frac{x_0(\alpha-3)+3-\alpha}{3(1-x_0)}.$$

We now inspect this derivative for the two values of $$\alpha$$ mentioned in the options.

$$g'(x_0)=\frac{x_0(2-3)+3-2}{3(1-x_0)} =\frac{-x_0+1}{3(1-x_0)}=\frac{1}{3}.$$
Hence $$L=|g'(x_0)|=\dfrac13.$$
Options A and B claim the limit is 0 or 1, so both are incorrect.

$$g'(x_0)=\frac{x_0(3-3)+3-3}{3(1-x_0)}=0.$$
Thus $$L=|g'(x_0)|=0.$$
This matches Option C, while Option D (value $$\tfrac23$$) is false.

Therefore the correct statement is:
Option C — For $$\alpha=3$$, $$\displaystyle\lim_{x\to x_0}\left|\frac{g(x)+e^{x_0}}{x-x_0}\right|=0.$$