Let f : ℝ → ℝ be a differentiable function satisfying f(x+y) = f(x)...

Given $\lim_{x \to 0} \dfrac{f(x)}{x} = 1$. Since the limit exists and the denominator approaches $0$, the numerator must also approach $0$, so $f(0) = 0$.

Applying L'Hopital's rule, $\lim_{x \to 0} f'(x) = f'(0) = 1$.

Differentiating the given functional equation $f(x+y) = f(x) + f(y) + 3x^2y + 3xy^2$ partially with respect to $x$ (treating $y$ as a constant):
$f'(x+y) = f'(x) + 6xy + 3y^2$

Substituting $x = 0$:
$f'(y) = f'(0) + 0 + 3y^2 = 1 + 3y^2$
Thus, $f'(x) = 3x^2 + 1$.

Integrating with respect to $x$:
$f(x) = x^3 + x + C$
Since $f(0) = 0$, $C = 0$, so $f(x) = x^3 + x$.

We are given that $g(x)$ is the inverse function of $f(x)$.
For $y = 10$, we have $f(x) = 10 \Rightarrow x^3 + x = 10 \Rightarrow x = 2$.

The second derivative of the inverse function is given by:
$g''(y) = \dfrac{-f''(x)}{(f'(x))^3}$

We have $f'(x) = 3x^2 + 1 \Rightarrow f'(2) = 13$
And $f''(x) = 6x \Rightarrow f''(2) = 12$

Therefore, $g''(10) = \dfrac{-12}{(13)^3} = \dfrac{-12}{2197}$.

The value of $|2197 \times g''(10)| = \left|2197 \times \left(\dfrac{-12}{2197}\right)\right| = 12$.

Answer: $12$