Here's a roadmap to a solution; the details are too tedious for me to figure out.
First note that if $(x,y,z)\in\Bbb{Q}^3$ satisfies
$$\frac{x^3}{y^3+z^3}+\frac{y^3}{x^3+z^3}+\frac{z^3}{y^3+x^3}=1,\tag{1}$$
then so does $(kx,ky,kz)$ for every nonzero integer $k$, so it suffices to find the integral solutions to $(1)$.
Now suppose $(x,y,z)\in\Bbb{Z}$ is and integral solution to $(1)$. Then clearing denominators and rearranging a bit we see that
$$x^9+y^9+z^9+x^3y^3z^3=0.$$
Then setting $(X,Y,Z):=(x^3,y^3,z^3)$ we get the following cubic equation in $X$, $Y$ and $Z$,
$$X^3+Y^3+Z^3+XYZ=0,\tag{2}$$
and then using the answer to this question as suggested in the comments, we can transform this to an elliptic curve with short Weierstrass form
$$Y^2=X^3+aX+b,$$
where
$$a=\frac{3085015005}{614656}\qquad\text{ and }\qquad b=-\frac{55345728797073}{240945152}.$$
Then LMFDB tells me that the corresponding minimal Weierstrass equation is in fact
$$y^2+xy+y=x^3-4x+6,\tag{3}$$
and that this elliptic curve has rank only the following five affine rational points:
$$(1,-1),\quad(2,2),\quad(2,-5),\quad(9,23),\quad(9,-33).$$
Of course equation $(2)$ has the obvious affine solutions (i.e. with $Z=1$)
$$(1,-1),\quad (-1,1),\quad (-1,-1),$$
that do not correspond to solutions to $(1)$, as they have $x^3+z^3=0$ or $y^3+z^3=0$.
Then there are two more rational solutions to $(2)$ with $Z=1$, and they can be made explicit by determining the rational transformation between $(2)$ and $(3)$. If these $X$- and $Y$-values of these two rational solutions are not perfect cubes, then the original equation $(1)$ has no rational solutions, which seems likely.
