Let $$g(x)$$ be a linear function and $$f(x) = \begin{cases} g(x), & x \leq 0 \\ \left(\frac{1+x}{2+x}\right)^{1/x}, & x > 0 \end{cases}$$, is continuous at $$x = 0$$. If $$f'(1) = f(-1)$$, then the value of $$g(3)$$ is

Let $$g(x)$$ be linear, so write $$g(x)=ax+b$$.

For continuity of $$f(x)$$ at $$x=0$$ we need the right-hand limit of the second branch to equal $$g(0)$$.
Consider $$y(x)=\left(\dfrac{1+x}{2+x}\right)^{1/x}$$ for $$x\gt 0$$.

Take logarithm:
$$\ln y=\frac{1}{x}\ln\!\left(\frac{1+x}{2+x}\right)=\frac{\ln(1+x)-\ln(2+x)}{x}.$$

Using series $$\ln(1+z)=z-\dfrac{z^{2}}{2}+O(z^{3})$$ about $$z=0$$,
$$\ln(1+x)-\ln(2+x)=-\ln2+\frac{x}{2}+O(x^{2}).$$
Hence
$$\ln y=\frac{-\ln2}{x}+\frac{1}{2}+O(x).$$

As $$x\to 0^{+}$$, $$\dfrac{-\ln2}{x}\to -\infty$$, so $$y(x)\to 0.$$
Therefore $$\displaystyle\lim_{x\to0^{+}}f(x)=0,$$ which must equal $$g(0)=b.$$
Thus $$b=0$$ and $$g(x)=ax.$$

Next, compute $$f'(1)$$ from the second branch.
Again set $$y(x)=\left(\dfrac{1+x}{2+x}\right)^{1/x}.$$
With $$\ln y=u(x)=\dfrac{1}{x}\ln\!\left(\dfrac{1+x}{2+x}\right),$$ we have
$$\frac{y'}{y}=u'(x).$$

Write $$v(x)=\ln(1+x)-\ln(2+x)$$ so that $$u(x)=\dfrac{v(x)}{x}.$$ Then
$$v'(x)=\frac{1}{1+x}-\frac{1}{2+x}.$$

At $$x=1$$:
$$v(1)=\ln2-\ln3=\ln\!\left(\frac{2}{3}\right),$$
$$v'(1)=\frac{1}{2}-\frac{1}{3}=\frac{1}{6}.$$

Therefore
$$u'(1)=\frac{1\cdot v'(1)-v(1)}{1^{2}}=\frac{1}{6}-\ln\!\left(\frac{2}{3}\right).$$

Also $$y(1)=\left(\frac{2}{3}\right)^{1}= \frac{2}{3}.$$ Hence
$$f'(1)=y'(1)=y(1)\,u'(1)=\frac{2}{3}\Bigl(\frac{1}{6}-\ln\!\frac{2}{3}\Bigr).$$

The condition $$f'(1)=f(-1)$$ gives
$$\frac{2}{3}\Bigl(\frac{1}{6}-\ln\!\frac{2}{3}\Bigr)=g(-1)=a(-1)=-a.$$

Solve for $$a$$:
$$a=-\frac{2}{3}\Bigl(\frac{1}{6}-\ln\!\frac{2}{3}\Bigr)=-\frac{1}{9}+\frac{2}{3}\ln\!\frac{2}{3}.$$

Finally,
$$g(3)=3a=3\!\left(-\frac{1}{9}+\frac{2}{3}\ln\!\frac{2}{3}\right) =-\frac{1}{3}+2\ln\!\frac{2}{3} =\ln\!\frac{4}{9}-\frac{1}{3} =\ln\!\frac{4}{9e^{1/3}}.$$

Hence $$g(3)=\log_e\frac{4}{9e^{1/3}},$$ which corresponds to Option D.