= Solution
Let $D_Q=(Q)-(O)$ on $E'$. Compatibility of the divisor-class maps with pullback gives a function $h\in K(E)^\times$ such that
$$
D_{\widehat\phi(Q)}=\phi^*D_Q+\operatorname{div}(h).
$$
If $f_Q$ has divisor $mD_Q$, then
$$
\operatorname{div}((f_Q\circ\phi)h^m)=mD_{\widehat\phi(Q)}.
$$
Use these functions in the divisor-evaluation formula for the <Weil pairing>. Pullback and pushforward satisfy
$$
(f_Q\circ\phi)(D_P)=f_Q(\phi_*D_P),
$$
while the factor $h^m$ contributes an $m$th power and cancels from the pairing. The two evaluations are therefore identical, giving
$$
e_m(P,\widehat\phi(Q))=e_m(\phi(P),Q)
$$
for every $P\in E[m]$ and $Q\in E'[m]$.
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