Find f(x).

Upon observation, $f(0) = 0$ for the above functional equation. This clearly eliminates Choices A and E.

If $f(x) = 0$ , then we end up with $x = 0$ (which does not hold for all $x \in \mathbb{R}$ ). Choice D is also eliminated.

Returning to our original functional equation, we find that:

$f(x) = x + \int_{0}^{1} (xy^2 + x^{2}y) \cdot f(y) dy = [\int_{0}^{1} y f(y) dy] \cdot x^2 + [1 + \int_{0}^{1} y^2 f(y) dy] \cdot x = \boxed{Ax^2+Bx}$ (where $A,B$ are real arbitrary constants)

Substituting this quadratic back into the original functional equation now produces:

$Ax^2 + Bx = x + \int_{0}^{1} (xy^2 + x^2y)(Ay^2 + By) dy;$

or $Ax^2 + Bx = x + \int_{0}^{1} Axy^4 + (Ax^2 + Bx)y^3 + Bx^2y^2 dy;$

or $Ax^2 + Bx = x + [\frac{A}{5} xy^5 + (\frac{A}{4}x^2 + \frac{B}{4}x)y^4 + \frac{B}{3}x^2y^3 |_{0}^{1}];$

or $Ax^2 + Bx = x + \frac{A}{5} x + \frac{A}{4}x^2 + \frac{B}{4}x + \frac{B}{3}x^2;$

or $Ax^2 + Bx = \frac{3A+4B}{12} x^2 + \frac{20+4A+5B}{20}x.$

After matching coefficients with one another, we arrive at the $2x2$ linear system:

$9A - 4B = 0$

$-4A + 15B = 20$

which yields $A = \frac{80}{119}, B = \frac{180}{119}$ as the unique solution. This finally gives us $\boxed{f(x) =\frac{80}{119} x^2 + \frac{180}{119}x}$ (or Choice C) as the desired function.