This is given by the transformation between Euler and Lagrange coordinates:
\vec{r}=\vec{r}(t,\vec{R}).
It describes the motion of a fluid element that was at the initial time, t=0 at position \vec{R}. This mapping between \vec{r} and \vec{R} is invertible as long as there are no singularities like shock waves etc. Then you can express every quantity in both the Lagrange coordinates \vec{R} and Euler coordinates \vec{r}. E.g., the velocity of a fluid element is given by
\vec{v}(t,\vec{R})=\partial_t \vec{r}(t,\vec{R}).
You can as well take the velocity field as function of the Euler coordinates. Then the relation is
\vec{v}(t,\vec{R})=\vec{v}[t,\vec{r}(t,\vec{R})] \equiv \vec{v}(t,\vec{r}).
Then you have to distinguish between partial time derivatives (at fixed Euler coordinates \vec{r}) and "material time derivatives (time derivatives at fixed Lagrange coordinates \vec{R}. E.g., for the equations of motion (Euler perfect fluid or viscous Navier Stokes etc) you need the acceleration of a given fluid element, i.e., the material time derivative
\mathrm{D}_t \vec{v}=\frac{\partial}{\partial t} \vec{v}(t,\vec{r})+\frac{\partial \vec{r}(t,\vec{R})}{\partial t} \cdot \vec{\nabla}_{\vec{r}} \vec{v}(t,\vec{r})=\frac{\partial}{\partial t} \vec{v}(t,\vec{r})+(\vec{v} \cdot \vec{\nabla}_r) \vec{v}(t,\vec{r}).
If you have the velocity field in Euler coordinates, \vec{v}(t,\vec{r}), you can thus simply calculate the trajectories of the fluid elements by solving the set of coupled differential equations as an initial-value problem:
\frac{\mathrm{d} \vec{r}}{\mathrm{d} t}=\vec{v}(t,\vec{r}), \quad \vec{r}(t=0)=\vec{R}
This gives you the Euler coordinates in terms of the Lagrange coordinates, \vec{r}=\vec{r}(t,\vec{R}).