Let f: ℝ arrow ℝ be such that f(xy) = f(x)f(y) , for all x, y ∈ ℝ...

Given $f(xy) = f(x)f(y)$ for all $x, y \in \mathbb{R}$ and $f(0) \neq 0$.

Substituting $y = 0$, we get $f(0) = f(x)f(0)$.

Since $f(0) \neq 0$, dividing by $f(0)$ gives $f(x) = 1$ for all $x \in \mathbb{R}$.

The given integral equation is:

$x^2 g(x) = \int\limits_1^x (t^2 f(t) - tg(t))\,dt$

Substituting $f(t) = 1$:

$x^2 g(x) = \int\limits_1^x (t^2 - tg(t))\,dt$

Differentiating both sides with respect to $x$ using Leibniz's rule:

$2x g(x) + x^2 g'(x) = x^2 - xg(x)$

$x^2 g'(x) + 3xg(x) = x^2$

Dividing by $x$ (since $x \ge 1$):

$x g'(x) + 3g(x) = x$

$g'(x) + \dfrac{3}{x}g(x) = 1$

This is a linear differential equation. The integrating factor is:

$\text{IF} = e^{\int \dfrac{3}{x}\,dx} = e^{3\ln x} = x^3$

Multiplying the differential equation by $x^3$:

$x^3 g'(x) + 3x^2 g(x) = x^3$

$\dfrac{d}{dx}(x^3 g(x)) = x^3$

Integrating both sides with respect to $x$:

$x^3 g(x) = \dfrac{x^4}{4} + C$

From the integral equation, substituting $x = 1$ gives $1^2 g(1) = \int\limits_1^1 (t^2 - tg(t))\,dt = 0$, so $g(1) = 0$.

Substituting $x = 1$ and $g(1) = 0$ into the integrated equation:

$1^3(0) = \dfrac{1^4}{4} + C \Rightarrow C = -\dfrac{1}{4}$

Thus, the function $g(x)$ is given by:

$x^3 g(x) = \dfrac{x^4 - 1}{4}$

$g(x) = \dfrac{x^4 - 1}{4x^3}$

Substituting $x = 2$:

$g(2) = \dfrac{2^4 - 1}{4(2^3)} = \dfrac{16 - 1}{32} = \dfrac{15}{32}$

Answer: $\dfrac{15}{32}$