Solution
= Solution
Use $y$ on the target and $x$ on the source, so the map of coordinate <rings> sends $y$ to $x^4$. A <closed point> of the real affine line corresponds to a <maximal ideal> generated either by $y-a$, $a\in\mathbb R$, or by a monic irreducible quadratic $q(y)$. The <scheme-theoretic fibre> over that point has coordinate algebra
$$
B_p=\mathbb R[x]\otimes_{\mathbb R[y]}\kappa(p)
=\mathbb R[x]/(q_p(x^4)).
$$
Here $\kappa(p)$ is the <residue field>, and $q_p=y-a$ in the real-point case. The fibre is a scheme, so its nilpotents must be retained.
For $a>0$, put $r=a^{1/4}$. The factors $x-r$, $x+r$, $x^2+r^2$ are pairwise coprime, with the last one irreducible over $\mathbb R$. The <Chinese remainder theorem> identifies $B_p$ with $\mathbb R\times\mathbb R\times\mathbb C$, giving three <irreducible components>.
For $a<0$, put $r=(-a)^{1/4}$. The exact factorization is
$$
x^4+r^4=(x^2+\sqrt2rx+r^2)(x^2-\sqrt2rx+r^2).
$$
Each quadratic has discriminant $-2r^2<0$, and they are distinct and coprime. Thus $B_p\cong\mathbb C\times\mathbb C$, giving two <irreducible components>.
For $a=0$, $B_p=\mathbb R[x]/(x^4)$. It has one <prime ideal>, generated by the class of $x$, so one <irreducible component>. It is not the reduced point $\operatorname{Spec}\mathbb R$: its class of $x$ is a nonzero <nilpotent element> with nilpotency index four.
Finally write the irreducible quadratic as $q_p(y)=(y-z)(y-\overline z)$, with $z\notin\mathbb R$. Over $\mathbb C$, $q_p(x^4)$ has four roots above $z$ and four above $\overline z$. They are all simple, because $z\ne0$, and the two sets are disjoint. None is real. Conjugation pairs them into four distinct irreducible real quadratics. Hence $B_p\cong\mathbb C^4$ and this fibre has four <irreducible components>.
The <real fourth-power fibre classification> is:
|| Target closed-point type
|| Real coordinate algebra of the fibre
|| <Irreducible components>
| $y-a$, $a>0$
| $\mathbb R^2\times\mathbb C$
| $3$
| $y-a$, $a<0$
| $\mathbb C^2$
| $2$
| $y$
| $\mathbb R[x]/(x^4)$
| $1$
| Irreducible quadratic
| $\mathbb C^4$
| $4$
All fibres in each row are isomorphic as affine schemes over $\mathbb R$. The rows are distinct: their numbers of <irreducible components> already distinguish them, and the zero fibre is additionally nonreduced. Therefore
$$
\boxed{\text{exactly four real-scheme isomorphism classes}}.
$$
These are ordinary scheme components, not geometric components counted after complexifying the base. Counting only real roots would miss both the complex-residue factors and the nonreal <closed points>.