Denote by $E(a, b, c)$ the given expression. First, we find the maximum value of $E(a, b, c)$ under the assumption that $a + b - c$, $b + c - a$, $c + a - b$ are non-negative.
Since $a^2 + b^2 - ab \ge ab$, it follows that $a^3 + b^3 = (a + b)(a^2 + b^2 - ab) \ge ab(a + b)$, so $a^3 + b^3 + abc \ge ab(a + b + c)$. Analogously, we have $b^3 + c^3 + abc \ge bc(a + b + c)$ and $c^3 + a^3 + abc \ge ca(a + b + c)$. Therefore:
$$E(a, b, c) = \sum \frac{a+b-c}{a^3+b^3+abc} \le \sum \frac{a+b-c}{ab(a+b+c)} = \frac{1}{a+b+c} \cdot \sum \frac{a+b-c}{ab}$$
Since $\sum \frac{a+b-c}{ab} = \frac{2(ab+bc+ca)-(a^2+b^2+c^2)}{abc} \le \frac{2(ab+bc+ca)-(ab+bc+ca)}{abc}$ (due to the well-known inequality $a^2 + b^2 + c^2 \ge ab + bc + ca$), we infer that
$$E(a, b, c) \le \frac{1}{a+b+c} \cdot \frac{ab+bc+ca}{abc} = \frac{1}{a+b+c} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \le 1.$$
Consider now the case when at least one of the numbers $a + b - c$, $b + c - a$, $c + a - b$ is negative. Without loss of generality, suppose that $a + b - c < 0$. Then $c > a + b > |a - b|$, so both numbers $b + c - a$ and $c + a - b$ are positive. It follows that:
$$\begin{align*} \sum \frac{a+b-c}{a^3+b^3+abc} &\le \frac{b+c-a}{b^3+c^3+abc} + \frac{c+a-b}{c^3+a^3+abc} \\ &\le \frac{b+c-a}{bc(a+b+c)} + \frac{c+a-b}{ca(a+b+c)} \\ &= \frac{1}{a+b+c} \cdot \frac{ac+bc-(a-b)^2}{abc} < \frac{1}{a+b+c} \cdot \frac{ab+bc+ca}{abc} \\ &= \frac{1}{a+b+c} \left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right) \le 1. \end{align*}$$
In summary, the maximum value of $E(a, b, c)$ is equal to $1$ and it is attained if and only if $a = b = c = 1$.