$$(x + y + z)^3 = x^3 + y^3 + z^3 + 3(x^2y + xy^2 + x^2z + y^2z + y^2x + xyz) + 6xyz,$$
while
$$(3(x + y + z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) xyz = 3(x + y + z)(xy + xz + yz) \\ = 3(x^2y + xy^2 + x^2z + xz^2 + y^2z + yz^2) + 9xyz.$$
Thus
$$(x + y + z)^3 - 3(x + y + z) \left(\frac{1}{x} + \frac{1}{y} + \frac{1}{z}\right) xyz = x^3 + y^3 + z^3 - 3xyz.$$
By assumptions, $x + y + z = 4$ and $\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{3}$. Hence $64 - 4xyz = x^3 + y^3 + z^3 - 3xyz$, implying $x^3 + y^3 + z^3 + xyz = 64$. Consequently, the expression $x^3 + y^3 + z^3 + xyz$ has only one value 64.