Let $f=3t$ then the cubic is:
$$
p(x) = (x+y)^3 - x^2 z + t x z (3 x + 3 y + t^2 z) = \prod (x-x_k)
$$
Let $\omega$ be a primitive cube root of unity, and $q(x) = x^3 - u x^2 + v x - w$ the cubic with roots $\sqrt[3]{x_k}$, then:
$$
\begin{align}
q(x)\,q(\omega x)\,q(\omega^2x) &= \prod (x-\sqrt[3]{x_k})(\omega x-\sqrt[3]{x_k})(\omega^2x-\sqrt[3]{x_k})
\\ &= \prod (x^3 - x_k)
\\ &= p(x^3)
\end{align}
$$
It follows that:
$$
\begin{align}
p(x^3) &= (x^3 - u x^2 + v x - w)(x^3 - \omega^2\,u x^2 + \omega\, v x - w)(x^3 - \omega\,u x^2 + \omega^2\,v x - w)
\\ &= \color{blue}{x^9} - x^6 (u^3 - 3 u v + \color{blue}{3 w}) + x^3 (-3 u v w + v^3 + \color{blue}{3 w^2}) - \color{blue}{w^3}
\\ &= \color{blue}{(x^3-w)^3} - x^6 u^3 + x^3 \left(3 u v x^3 - 3u v w + v^3\right)
\end{align}
$$
By inspection, comparing to $\,p(x^3) = (x^3+y)^3 - x^6 z + x^3 (3 t z x^3 + 3 t z y + t^3 z^2)\,$:
$$
\begin{cases}
\begin{align}
u &= \sqrt[3]{z}
\\ v &= t \sqrt[3]{z^2}
\\ w &= -y
\end{align}
\end{cases}
$$
Therefore $\,q(x)=x^3 - \sqrt[3]{z}\,x^2 + t\sqrt[3]{z^2}\,x + y\,$, so $\,\sqrt[3]{x_1}+\sqrt[3]{x_2}+\sqrt[3]{x_3} = u = \sqrt[3]{z}\,$.
(Note: just for reference, below is the first and very different revision of my answer.)
To get this out the way while looking forward to a more satisfying answer, the following is a brute force breakdown. With $f=3t$ the cubic is:
$$(x+y)^3 - x^2 z + t x z ( 3 x + 3 y + t^2 z) \;=\; x^3 + x^2 (3 y + 3 t z - z) + x (3 y^2 + 3 t y z + t^3 z^2) + y^3$$
@River Li's answer here provides a formula for $\,\sum \sqrt[3]{x_k}\,$ for generic cubics with distinct real roots (the assumption of rational coefficients is not needed or used AFAICT).
In general, suppose $x_1,x_2,x_3$ are the three distinct real roots of $x^3+bx^2+cx+d^3=0$,
where $b,c,d$ are rational numbers. We have
\begin{equation}
\sqrt[3]{x_1}+\sqrt[3]{x_2}+\sqrt[3]{x_3} = \sqrt[3]
{-\frac{3}{2}\Big(\sqrt[3]{Q+4\sqrt{\Delta}}+\sqrt[3]{Q-4\sqrt{\Delta}}\Big) -b-6d } \tag{*}
\end{equation}
where $Q = 24bd^2+36d^3+4bc+24cd, \Delta = -4b^3d^3-27d^6+18bcd^3+b^2c^2-4c^3$.
Here, $\Delta$ is the discriminant of $x^3+bx^2+cx+d^3=0$.
With those notations, the cubic here has:
$$
\begin{align}
b &= 3 y + 3 t z - z
\\ c &= 3 y^2 + 3 t y z + t^3 z^2
\\ d &= y
\\ Q &= 4 (t z + 3 y) (3 t^3 z^2 - t^2 z^2 + 12 t y z + 18 y^2 - 3 y z)
\\ \Delta &= -z^2 ((t - 1) t^2 z- y)^2 (t^2 (4 t - 1) z^2 + 2 (9 t - 2) y z + 27 y^2)
\\ Q^2 - 16 \Delta &= \big(4 (t^3 z^2 + (6 t - 1) y z + 9 y^2)\big)^3
\end{align}
$$
With the notations from my answer here about denesting cube roots like $\,\sqrt[3]{Q\pm4\sqrt{\Delta}}\,$:
$$
\begin{align}
r &= 4 (t^3 z^2 + (6 t - 1) y z + 9 y^2)
\\ s &= - t z - 3 y
\\ \sqrt[3]{Q+4\sqrt{\Delta}} + \sqrt[3]{Q-4\sqrt{\Delta}} = 2s &= -2(t z + 3 y)
\end{align}
$$
It follows that:
$$
{-\frac{3}{2}\Big(\sqrt[3]{Q+4\sqrt{\Delta}}+\sqrt[3]{Q-4\sqrt{\Delta}}\Big) -b-6d} = 3(t z + 3 y) - (3 y + 3 t z - z) -6y = z
$$
Which by $\,(*)\,$ concludes that $\,\sqrt[3]{x_1}+\sqrt[3]{x_2}+\sqrt[3]{x_3} = \sqrt[3]{z}\,$.
