New Day Calculus Functions.

Let us differentiate the above functional equation with respect to x & y, which yields:

$f'(x+y) = f'(x) + 2018y$ (i)

$f'(x+y) = f'(y) + 2018x$ (ii)

and equating (i) with (ii) produces $f'(x) = 2018x + f'(y) - 2018y \Rightarrow f'(x) = 2018x + A (A \in \mathbb{R})$ (iii). Integrating (iii) with respect to x next gives:

$f(x) = 1009x^2 + Ax + B$ (iv)

We need two initial conditions to solve for (iv), and if we take $x = y = 0$ into the original functional equation, we obtain:

$f(0+0) = f(0) + f(0) + 2018 \cdot 0^2 \Rightarrow f(0) = 0$

Coupling this condition with $f(1) = 1009$ yields $A = B = 0$ , which ultimately produces $\boxed{f(x) = 1009x^2}.$ Thus, $2018^2 \cdot f(\frac{1}{2018}) = 2018^2 \cdot 1009 \cdot \frac{1}{2018^2} = \boxed{1009}$ is the final result.