Answer: $c=1$.
If $x = y = z = \sqrt[6]{4c}$ then we get
$$\frac{x^3 y + y^3 z + z^3 x}{x + y + z} + \frac{4c}{xyz} = 4\sqrt{c} \ge 2c + 2$$
and hence $(\sqrt{c}-1)^2 \le 0$. Therefore, all $c \ne 1$ do not satisfy the inequality. Let us show that the inequality holds for $c = 1$. By AM-GM inequality we have

$$x^3y + \frac{4}{xy} \geq 4x$$
$$y^3z + \frac{4}{yz} \geq 4y$$
$$z^3x + \frac{4}{zx} \geq 4z$$
Side by side summation of these inequalities yields
$$x^3y + y^3z + z^3x + \frac{4(x + y + z)}{xyz} \geq 4(x + y + z).$$
Therefore,
$$\frac{x^3y + y^3z + z^3x}{x + y + z} + \frac{4}{xyz} \geq 4$$
and we are done.