= Solution
A morphism $\phi:\mathbb P^2\to\mathbb P^1$ is represented by homogeneous forms $F,G$ of the same degree with no common projective zero. If that degree were positive, the two plane curves $F=0$ and $G=0$ would intersect, producing a base point. Thus the degree is zero and $\phi$ is constant.
If a closed subvariety of $(\mathbb P^1)^3$ were isomorphic to $\mathbb P^2$, composing its embedding with each coordinate projection would give three constant maps. Their product would be constant, contradicting that it is an embedding.
The Riemann--Hurwitz theorem states
$$
2g_X-2=d(2g_Y-2)+\sum_{P\in X}(e_P-1)
$$
for a degree-$d$ nonconstant map of smooth projective curves.
Project $X$ to the second $\mathbb P^1$. This is a degree-two map. Its quadratic discriminant in the $x$ variables is homogeneous of degree six in $y$, so smoothness gives six branch points counted with multiplicity. Hence
$$
2g_X-2=2(-2)+6=2,
$$
and $g_X=2$. A smooth plane curve has genus $(d-1)(d-2)/2$, which is never two for an integer $d$, so $X$ is not isomorphic to a smooth plane curve.
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