1. Given piecewise function

$$f(x)= \begin{cases} g(x), & x\le 0 \\ \left(\dfrac{1+x}{2+x}\right)^{1/x}, & x>0 \end{cases}$$

where $g(x)$ is linear, and $f(x)$ is continuous at $x=0$.

We also have

$f'(1)=f(-1).$

We need to find $g(3)$.

---

2. Use continuity at $x=0$ to determine $g(0)$

Since $f$ is continuous at $x=0$,

$g(0)=\lim_{x\to 0^+}\left(\frac{1+x}{2+x}\right)^{1/x}.$

Let

$L=\lim_{x\to 0^+}\left(\frac{1+x}{2+x}\right)^{1/x}.$

Take logarithm:

$\ln L=\lim_{x\to 0^+}\frac{1}{x}\ln\left(\frac{1+x}{2+x}\right).$

Now,

$\ln\left(\frac{1+x}{2+x}\right)=\ln(1+x)-\ln(2+x).$

As $x\to 0^+$,

$\ln(1+x)\to 0, \qquad \ln(2+x)\to \ln 2.$

Hence

$\ln\left(\frac{1+x}{2+x}\right)\to -\ln 2 \neq 0.$

Therefore,

$\frac{1}{x}\ln\left(\frac{1+x}{2+x}\right)\to -\infty,$

so

$L=e^{-\infty}=0.$

Thus,

$g(0)=0.$

Since $g(x)$ is linear, let

$g(x)=mx+c.$

From $g(0)=0$, we get $c=0$, so

$g(x)=mx.$

---

3. Compute $f(-1)$

Since $-1\le 0$,

$f(-1)=g(-1)=-m.$

---

4. Compute $f'(1)$ for the right branch

For $x>0$,

$f(x)=\left(\frac{1+x}{2+x}\right)^{1/x}.$

Take logarithm:

$\ln f(x)=\frac{1}{x}\ln\left(\frac{1+x}{2+x}\right).$

Let

$y=f(x), \qquad \ln y=\frac{1}{x}\ln\left(\frac{1+x}{2+x}\right).$

Differentiate:

$\frac{y'}{y}=\frac{d}{dx}\left[\frac{1}{x}\ln\left(\frac{1+x}{2+x}\right)\right].$

Using product/quotient rule,

$\frac{y'}{y}=-\frac{1}{x^2}\ln\left(\frac{1+x}{2+x}\right)+\frac{1}{x}\cdot \frac{d}{dx}\left[\ln\left(\frac{1+x}{2+x}\right)\right].$

Now,

\frac{d}{dx}\ln\left(\frac{1+x}{2+x}\right)=\frac{1}{1+x}-\frac{1}{2+x}=rac{1}{(1+x)(2+x)}.

So,

$\frac{y'}{y}=-\frac{1}{x^2}\ln\left(\frac{1+x}{2+x}\right)+\frac{1}{x(1+x)(2+x)}.$

At $x=1$,

$f(1)=\left(\frac{2}{3}\right)^1=\frac{2}{3}.$

Also,

$$=-\ln\left(\frac{2}{3}\right)+\frac{1}{6}.$$

Hence,

$f'(1)=\frac{2}{3}\left[-\ln\left(\frac{2}{3}\right)+\frac{1}{6}\right].$

Since $-\ln(2/3)=\ln(3/2)$,

$f'(1)=\frac{2}{3}\ln\left(\frac{3}{2}\right)+\frac{1}{9}.$

---

5. Use the condition $f'(1)=f(-1)$

Given

$f'(1)=f(-1),$

so

$\frac{2}{3}\ln\left(\frac{3}{2}\right)+\frac{1}{9}=-m.$

Thus,

$$$$

Since $g(x)=mx$,

$g(3)=3m=-2\ln\left(\frac{3}{2}\right)-\frac{1}{3}.$

Now simplify:

$-2\ln\left(\frac{3}{2}\right)=\ln\left(\frac{2}{3}\right)^2=\ln\left(\frac{4}{9}\right).$

Therefore,

$g(3)=\ln\left(\frac{4}{9}\right)-\frac{1}{3}.$

Also,

$\ln\left(\frac{4}{9e^{1/3}}\right)=\ln\left(\frac{4}{9}\right)-\frac{1}{3}.$

Hence,

$g(3)=\ln\left(\frac{4}{9e^{1/3}}\right).$

---

6. Compare with options

This matches Option C:

$\boxed{\ln\left(\frac{4}{9e^{1/3}}\right)}.$

---

7. Verification with stored answer

Stored correct answer: C.

Our derived answer is also C. So they agree.