The correct answer is C).\n\nLet the given expression be $E$.\nWe are given $x, y, z > 0$ and $x+y+z=1$.\n\nFirst, let's simplify the denominators of each term in the expression $E$ using the condition $x+y+z=1$:\n\nFor the first term: $2x+y+z = x + (x+y+z) = x+1$.\nFor the second term: $x+2y+z = y + (x+y+z) = y+1$.\nFor the third term: $x+y+2z = z + (x+y+z) = z+1$.\n\nSo, the expression $E$ can be rewritten as:\n$$ E = \\frac{x^2}{x+1} + \\frac{y^2}{y+1} + \\frac{z^2}{z+1} $$\n\nNow, let's simplify each term. Consider a generic term $\\frac{t^2}{t+1}$. We can perform algebraic manipulation:\n$$ \\frac{t^2}{t+1} = \\frac{t^2-1+1}{t+1} = \\frac{(t-1)(t+1)+1}{t+1} = (t-1) + \\frac{1}{t+1} $$\n\nApplying this simplification to each term in $E$:\n$$ E = \\left(x-1+\\frac{1}{x+1}\\right) + \\left(y-1+\\frac{1}{y+1}\\right) + \\left(z-1+\\frac{1}{z+1}\\right) $$\n\nGroup the terms:\n$$ E = (x+y+z) - (1+1+1) + \\left(\\frac{1}{x+1} + \\frac{1}{y+1} + \\frac{1}{z+1}\\right) $$\n\nSubstitute $x+y+z=1$:\n$$ E = 1 - 3 + \\left(\\frac{1}{x+1} + \\frac{1}{y+1} + \\frac{1}{z+1}\\right) $$\n$$ E = -2 + \\left(\\frac{1}{x+1} + \\frac{1}{y+1} + \\frac{1}{z+1}\\right) $$\n\nTo maximize $E$, we need to maximize the sum $S = \\frac{1}{x+1} + \\frac{1}{y+1} + \\frac{1}{z+1}$.\n\nLet's analyze the function $f(t) = \\frac{1}{t+1}$ for $t>0$.\nFirst derivative: $f'(t) = -\\frac{1}{(t+1)^2}$\nSecond derivative: $f''(t) = \\frac{2}{(t+1)^3}$\n\nSince $t>0$, $f''(t) > 0$, which means $f(t)$ is a convex function.\n\nFor a convex function $f(t)$, given a constraint $\sum t_i = K$ (constant), the sum $\sum f(t_i)$ is minimized when all $t_i$ are equal, and maximized when the $t_i$ values are at the 'extremes' of their domain. In this case, the domain for $x,y,z$ is $(0,1)$ such that $x+y+z=1$. The 'extremes' occur when one variable approaches 1 and the other two approach 0.\n\nLet's consider the scenario where one variable approaches 1 and the others approach 0. For example, let $x \\to 1$, $y \\to 0^+$, and $z \\to 0^+$. (Note: Since $x,y,z$ must be positive, we consider the limit as they approach these boundary values).\n\nAs $x \\to 1, y \\to 0^+, z \\to 0^+$:\n* $\\frac{1}{x+1} \\to \\frac{1}{1+1} = \\frac{1}{2}$\n* $\\frac{1}{y+1} \\to \\frac{1}{0+1} = 1$\n* $\\frac{1}{z+1} \\to \\frac{1}{0+1} = 1$\n\nSo, the sum $S = \\frac{1}{x+1} + \\frac{1}{y+1} + \\frac{1}{z+1}$ approaches $ \\frac{1}{2} + 1 + 1 = \\frac{5}{2} $. \n\nTherefore, the maximum value of $E$ is approached as:\n$$ E = -2 + \\frac{5}{2} = \\frac{1}{2} $$\n\nTo provide context for other options:\nIf $x=y=z=1/3$ (where the sum of a convex function is minimized):\n$S = \\frac{1}{1/3+1} + \\frac{1}{1/3+1} + \\frac{1}{1/3+1} = 3 \\times \\frac{1}{4/3} = 3 \\times \\frac{3}{4} = \\frac{9}{4}$.\nIn this case, $E = -2 + \\frac{9}{4} = \\frac{1}{4}$. This is the minimum value of $E$, corresponding to option A.\n\nSince the question asks for the maximum value, we choose C) $\\frac{1}{2}$.\n\nThe final answer is $\\boxed{\\frac{1}{2}}$.