= Solution
For $E:y^2=x^3+1$, let $\alpha(x,y)=(\zeta x,y)$. The three points $P,\alpha P,\alpha^2P$ lie on the horizontal line through $P$, so their sum is zero. Thus
$$
\alpha^2+\alpha+1=0
$$
in the <endomorphism ring of an elliptic curve>. Since $\deg\alpha=1$ and complex conjugation sends $\alpha$ to $\alpha^2$, degree on $\mathbb Z[\alpha]$ is the <Eisenstein-integer norm>:
$$
\deg(m+n\alpha)
=(m+n\alpha)(m+n\alpha^2)
=m^2-mn+n^2.
$$
A separable isogeny is determined by its kernel up to unique isomorphism of its target, and every finite Galois-stable subgroup of an elliptic curve is the kernel of the corresponding quotient isogeny. Put $T=(-1,0)$. The three nonzero points of $E[2]$ are $T,\alpha T,\alpha^2T$, and
$$
(\alpha-\alpha^2)T=\alpha T+\alpha^2T=T.
$$
Hence $\varphi(\alpha-\alpha^2)$ kills $\ker\varphi=\{O,T\}$ and factors uniquely through $\varphi$:
$$
\varphi(\alpha-\alpha^2)=\psi\varphi
$$
for an isogeny $\psi:E'\to E'$. Degrees give
$$
\deg\psi=\deg(\alpha-\alpha^2)=3.
$$
Moreover,
$$
(\alpha-\alpha^2)^2=\alpha+alpha^2-2=-3.
$$
Composing the factorization twice and using the surjectivity of $\varphi$ gives $\psi^2=[-3]$.
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