If a differentiable function satisfies
$(x - y)f(x + y) - (x + y)f(x - y) = 2(x^2y - y^3) \, \forall x, y \in \mathbb{R}$ and $f(1) = 2$, then

Let $x + y = u$ and $x - y = v$.

Then $x = \dfrac{u+v}{2}$ and $y = \dfrac{u-v}{2}$.

The right hand side of the given equation can be simplified as:

$2(x^2y - y^3) = 2y(x^2 - y^2) = 2y(x-y)(x+y) = 2 \left( \dfrac{u-v}{2} \right) (v) (u) = uv(u-v)$

Substituting these into the given equation, we get:

$v f(u) - u f(v) = uv(u-v)$

Dividing the entire equation by $uv$ (for $u, v \neq 0$):

$\dfrac{f(u)}{u} - \dfrac{f(v)}{v} = u - v$

Rearranging the terms to separate the variables:

$\dfrac{f(u)}{u} - u = \dfrac{f(v)}{v} - v$

Since this holds for all $u, v \neq 0$, both sides must be equal to a constant $c$:

$\dfrac{f(u)}{u} - u = c \Rightarrow f(u) = u^2 + cu$

Since $f(x)$ is a differentiable function, it must be continuous everywhere, including at $x = 0$.

$f(0) = \lim_{u \to 0} (u^2 + cu) = 0$

Thus, $f(x) = x^2 + cx$ for all $x \in \mathbb{R}$.

We are given $f(1) = 2$. Substituting $x = 1$:

$1^2 + c(1) = 2 \Rightarrow c = 1$

Therefore, the function is $f(x) = x^2 + x$.

Checking the given options:

$\bullet$ $f(x) = x^2 + x$ is a polynomial function.

$\bullet$ $f(3) = 3^2 + 3 = 12$.

$\bullet$ $f(0) = 0^2 + 0 = 0$.

Answer: $f(x)$ must be a polynomial function; $f(3) = 12$; $f(0) = 0$