= Solution
Use the corrected interpolant from part (a). The error has $e(x_i)=e'(x_i)=0$, and the chosen auxiliary function has $\phi(x_i)=\phi'(x_i)=0$ as well as $\phi(x)=0$. There are $n+2$ distinct zeros of $\phi$: all the nodes and the additional evaluation point. <Rolle's theorem> supplies $n+1$ zeros of $\phi'$ between these ordered zeros. Together with its $n+1$ nodal zeros, this gives $2n+2$ distinct derivative zeros. Repeated Rolle applications therefore give an interior $\xi$ such that $\phi^{(2n+2)}(\xi)=0$.
\b[The extra claim that $x$ itself must be a zero of $\phi'$ is false.] For nodes zero and one and $f(t)=t^5$, the Hermite interpolant is $3t^3-2t^2$, so $e(t)=t^2(t-1)^2(t+2)$. The auxiliary function is then $\phi(t)=t^2(t-1)^2(t-x)$, and $\phi'(x)=x^2(x-1)^2\ne0$ at any interior evaluation point. The corrected zero count above is all that is needed.
Since $h$ has degree at most $2n+1$ and $\pi^2$ is monic of degree $2n+2$, differentiating the auxiliary function at the supplied $\xi$ gives
$$
0=f^{(2n+2)}(\xi)-\frac{e(x)}{\pi(x)^2}(2n+2)!.
$$
Hence
$$
\boxed{e(x)=\pi(x)^2\frac{f^{(2n+2)}(\xi)}{(2n+2)!}}.
$$
At interpolation nodes the error and the prefactor are zero, so the formula remains valid there as well.
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