I am looking for a general solution, but here is an specific example.
The Clebsch Diagonal Cubic is given by this equation in homogeneous coordinates
$$X^3 + Y^3 + Z^3 + W^3 = (X+Y+Z+W)^3$$
Apparently, by "taking a plane at infinity" one can obtain an equivalent implicit equation in non-homogeneous Cartesian $x,y,z$ coordinates:
$$81(x^3+y^3+z^3)-189(x^2y+x^2z+y^2x+y^2z+z^2x+z^2y)+54xyz+126(xy+xz+yz)-9(x^2+y^2+z^2)-9(x+y+z)+1=0$$
This is the form I want to work in.
Given an equation in $X,Y,Z,W$, how can I obtain an equation in $x,y,z$?
Essentially, when you go from projective space to euclidean space you pick a hyperplane, mark it as the hyperplane at infinity, and "remove" it (and all points it contains) from consideration. This is to signal the fact that a plane at "infinity" does not exist in euclidean space. The choice of hyperplane at infinity is arbitrary. In this case, the chosen hyperplane is $x_0 + x_1 + x_2 + \frac{x_3}{2} = 0$.
I'll illustrate the procedure in 3D. Consider the projective plane $ax_1 + bx_2 + cx_3 + dx_4= 0$. Let's pick the "plane at infinity" as $x_4 = 0$. To remove it we interpret $\frac{x_1}{x_4} = x$ and $\frac{x_2}{x_4} = y$ and $\frac{x_3}{x_4} = z$ as $x_4$ can never be 0 under this euclidean specialization. Then the equation of the plane under the given euclidean specialization is $ax + by + cz + d = 0$.
We could have chosen $x_2 = 0$ as the "plane at infinity" and obtained $ax + b + cy + dz = 0$ as the plane under this euclidean specialization. So the equation you obtain will depend on the hyperplane you choose as the "hyperplane at infinity".