= Solution
Use the polynomial graded <ring> $R[x_0,x_1,x_2,x_3]$, as in the PDF; the doubled brackets in the converted TeX are an OCR error. Let $I$ denote the homogeneous defining <ideal>. The four cubic coordinates in the proposed map never vanish simultaneously on $\mathbb P_F^1$, since $y_0^3$ and $y_1^3$ already have that property. Substitution makes each generator of $I$ zero, so they define a morphism into the generic fibre.
The two standard opens $D_+(x_0)$ and $D_+(x_3)$ cover $X$. Indeed, a homogeneous prime containing $x_0,x_3$ contains $x_2$ by $x_0x_3^2=x_2^3$, and then contains $x_1$ by $x_1^2=\pi^2x_0x_2$. Such a prime contains the irrelevant <ideal> and gives no point of Proj.
On $D_+(x_0)$ put $a=x_1/x_0$, $b=x_2/x_0$, $c=x_3/x_0$. Over $F$, where $\pi$ is invertible, the equations give
$$
a=\pi t,\qquad b=t^2,\qquad c=t^3,\qquad t=a/\pi.
$$
Thus the generic-fibre chart is exactly $\operatorname{Spec}F[t]$. The proposed map restricts to this <isomorphism> with $t=y_1/y_0$. On $D_+(x_3)$ put $s=x_2/x_3$. The equations give
$$
x_0/x_3=s^3,\qquad x_1/x_3=\pi s^2,
$$
so this chart is $\operatorname{Spec}F[s]$, and the map has $s=y_0/y_1$. On the overlap $s=t^{-1}$. These are the standard two charts of the <projective line>, proving the scheme-level <isomorphism>
$$
\boxed{\mathbb P_F^1\xrightarrow{\sim}X_F.}
$$
Equivalently, rescaling $x_1$ by $\pi^{-1}$ gives the usual <twisted cubic>. The chart proof also shows that the extra cubic relation introduces no additional generic-fibre <scheme> structure.
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