Let f: ℝ → ℝ and g: ℝ → ℝ be functions satisfying f(x + y) = f(x) +...

Given $f\left(x + y\right) = f\left(x\right) + f\left(y\right) + f\left(x\right)f\left(y\right)$
Put $x = y = 0$ in given relation.

$⇒ f\left(0\right) = f\left(0\right) + f\left(0\right) + {f}^{2}\left(0\right)$

$⇒ f\left(0\right) = 0$ or $- 1$

$∵ f\left(x + y\right) = f\left(x\right) + f\left(y\right) + f\left(x\right) \cdot f\left(y\right)$

$⇒ \dfrac{f\left(x + y\right) - f\left(x\right)}{y} = \dfrac{f\left(y\right)\left(1 + f\left(x\right)\right)}{y}$

$⇒ \lim_{y \to 0}\left(\dfrac{f ( x + y ) - f ( x )}{y}\right) = \lim_{y \to 0}\left(1 + f\left(x\right)\right) \cdot \dfrac{f\left(y\right)}{y}$

$∵ \lim_{x \to 0}g\left(x\right) = \lim_{x \to 0}\dfrac{f\left(x\right)}{x} = 1$

$⇒ {f}^{'}\left(x\right) = 1 + f\left(x\right)$

$⇒ {f}^{'}\left(0\right) = 1 + f\left(0\right)$

$⇒ {f}^{'}\left(0\right) = 1 + 0$

$⇒ {f}^{'}\left(0\right) = 1$

Again $\dfrac{{f}^{'}\left(x\right)}{1 + f\left(x\right)} = 1 ⇒ ∫ \dfrac{{f}^{'}\left(x\right)dx}{1 + f\left(x\right)}dx = ∫ dx$

$⇒ \ln \left(1 + f\left(x\right)\right) = x + C$

⇒ \ln \left[1 + f\left(x\right)\right] = x$\left(∵ C = 0\right)$

$⇒ 1 + f\left(x\right) = {e}^{x}$

$⇒ f\left(x\right) = {e}^{x} - 1 ⇒ {f}^{'}\left(x\right) = {e}^{x}$

$⇒ {f}^{'}\left(1\right) = e$

Also, $f\left(x\right)$ is differentiable for every $x \in R$.

$g\left(x\right) = \dfrac{f\left(x\right)}{x} = \dfrac{{e}^{x} - 1}{x}{y}^{'}\left({0}^{+}\right) = \lim_{h \to 0}\dfrac{g\left(0 + h\right) - g\left(0\right)}{h}$

If $g\left(0\right) = 1$ then ${g}^{'}\left({0}^{+}\right) = \lim_{h \to 0}\dfrac{\dfrac{{e}^{h} - 1}{h} - 1}{h} = \lim_{h \to 0}\dfrac{{e}^{h} - 1 - h}{{h}^{2}} = \dfrac{1}{2}$

${g}^{'}\left({0}^{-}\right) = \lim_{h \to 0}\dfrac{g\left(0 - h\right) - g\left(0\right)}{- h} = \lim_{h \to 0}\dfrac{{e}^{- h} - 1}{\dfrac{- h}{- h}}$

$\lim_{h \to 0}\dfrac{{e}^{- h} - 1 + h}{{h}^{2}} = \dfrac{1}{2}$

$g\left(x\right)$ is differentiable for every $x \in R$.