Let $$f:R \rightarrow R$$ and $$g:R \rightarrow R$$ be functions satisfying
$$f(x + y)=f(x)+f(y)+f(x)f(y)$$ and $$f(x) = xg(x)$$
for all $$𝑥, 𝑦 \in R$$. If $$\lim_{x \rightarrow 0} g(x) = 1$$, then which of the following statements is/are TRUE?

Define a new function $$h:\mathbb{R}\rightarrow\mathbb{R}$$ by $$h(x)=1+f(x)\;.$$

Using the given relation $$f(x+y)=f(x)+f(y)+f(x)f(y)$$ we get

$$h(x+y)-1=\bigl(h(x)-1\bigr)+\bigl(h(y)-1\bigr)+\bigl(h(x)-1\bigr)\bigl(h(y)-1\bigr)$$
$$\Longrightarrow\;h(x+y)-1=h(x)h(y)-1$$
$$\Longrightarrow\;\boxed{\,h(x+y)=h(x)\,h(y)\,}\;.$$

Also $$h(x)=1+f(x)=1+xg(x)\;.$$ Since $$\displaystyle\lim_{x\to 0}g(x)=1$$, we have

$$\lim_{x\to 0}h(x)=\lim_{x\to 0}\bigl(1+xg(x)\bigr)=1\;,$$

so $$h$$ is continuous at $$x=0$$.

Standard theory for the Cauchy exponential equation states that any function satisfying $$h(x+y)=h(x)h(y)$$ and continuous at $$0$$ must be of the form

$$\boxed{\,h(x)=e^{kx}\,}\qquad(k\in\mathbb{R}).$$

Therefore

$$f(x)=h(x)-1=e^{kx}-1\;.$$

Using $$f(x)=xg(x)$$ for $$x\neq 0$$,

$$g(x)=\frac{e^{kx}-1}{x}\qquad(x\neq 0).$$

Taking the limit as $$x\to 0$$,

$$\lim_{x\to 0}g(x)=\lim_{x\to 0}\frac{e^{kx}-1}{x}=k.$$

This limit is given to be $$1$$, hence $$k=1$$. Thus

$$\boxed{\,f(x)=e^{x}-1},\qquad\boxed{\,g(x)=\frac{e^{x}-1}{x}}\;(x\neq 0),\qquad g(0)=1.$

Verification of the statements

Statement A. Since $$f(x)=e^{x}-1$$, $$f$$ is the difference of two functions that are differentiable everywhere. Hence $$f$$ is differentiable for every $$x$$\in$$$$\mathbb{R}$$$$. True.

Statement B. With $$g(0)=1$$ the completed definition of $$g$$ is

$$g(x)=$$\begin{cases}$$\dfrac{e^{x}-1}{x}, & x$$\neq$$ 0,\\[6pt]1,&x=0.\end{cases}$$

For $$x$$\neq$$ 0$$, $$g$$ is a quotient of differentiable functions, hence differentiable. At $$x=0$$, compute the derivative:

$$g'(0)=$$\lim_{x\to$$ 0}$$\frac{g(x)-g(0)}{x-0}$$=% $$\lim_{x\to$$ 0}$$\frac{\dfrac{e^{x}$$-1}{x}-1}{x}% =$$\lim_{x\to$$ 0}$$\frac{e^{x}$$-1-x}{x^{2}}% =$$\frac{1}{2}$$,$$

which exists. Therefore $$g$$ is differentiable at every real $$x$$. True.

Statement C. $$f'(x)=e^{x}$$, hence $$f'(1)=e$$\neq$$ 1$$. False.

Statement D. $$f'(0)=e^{0}=1$$. True.

Thus the correct assertions are:

Option A (f is differentiable everywhere), Option B (g is differentiable everywhere when $$g(0)=1$$), and Option D ($$f'(0)=1$$).

Answer: Option A, Option B, Option D.