It’s easy but...

Note that $g(x) := f(x) - x^2 \implies g(x+y) = g(x) + g(y) \iff\!\!\!^{\color{#3D99F6}[1]}\,\,\, g(x) = C\cdot x \quad \text{ for some } C$ so that $f(x) = x^2 + C\cdot x$ for some $C.$

Now set $x=10$ : $135 = f(10) = 10^2 + 10C \implies C = 3.5$ so that $f\left(\tfrac{1}{2}\right) = \left(\tfrac{1}{2}\right)^2 + 3.5\left(\tfrac{1}{2}\right) = \boxed{2}$

---

Remark on $^{\color{#3D99F6}[1]}$ : This step (specifically the "only if" direction) actually assumes $g$ (and therefore $f$ ) is continuous at zero. This isn't necessarily the case, but if it bothers you, simply replace $f$ with $f|_\mathbb{Q}$ by restricting its domain to the rational numbers.

Let's try a calculus approach to this functional equation. Integrating it with respect to $x$ and $y$ respectively yields:

$f'(x+y) = f'(x) + 2y$ AND $f'(x+y) = f'(y) + 2x$

which if we equate these two above equations, we obtain:

$f'(x) + 2y = f'(y) + 2x \Rightarrow f'(x) = 2x + [f'(y) - 2y] = 2x + A$ (for $A \in \mathbb{R}).$ (i).

Integrating (i) produces:

$f(x) = x^2 + Ax + B$ (ii).

Substituting (ii) back into the original functional equation now yields:

$(x+y)^2 + A(x+y) + B = (x^2 + Ax + B) + (y^2 + Ay + B) + 2xy \Rightarrow x^2 + 2xy + y^2 + A(x+y) + B = x^2 + 2xy + y^2 + A(x+y) + 2B$ (iii)

Upon observation, we must have $B = 0$ in order for (iii) to hold for all $x,y \in \mathbb{R}.$ Using the provided initial condition $f(10) = 135$ , we now solve for $A$ :

$135 = 10^2 + 10A \Rightarrow A = 3.5$

which leaves us with $f(x) = x^2 + 3.5x$ , and $f(\frac{1}{2}) = (\frac{1}{2})^2 + 3.5(\frac{1}{2}) = \boxed{2}.$