= Solution
Use the <Rabinowitsch trick>. In the polynomial ring $k[x_1,\ldots,x_n,t]$, consider
$$
I=(J,1-tf).
$$
If $I$ were proper, it would lie in a maximal ideal. Part (a), together with algebraic closedness of $k$, is the <Weak Hilbert Nullstellensatz>, so that maximal ideal would be evaluation at a point $(a_1,\ldots,a_n,b)\in k^{n+1}$. We would then have
$$
a=(a_1,\ldots,a_n)\in Z(J),
\qquad 1-bf(a)=0.
$$
But the hypothesis gives $f(a)=0$, a contradiction. Therefore $I$ is the unit ideal.
There are polynomials $g_j$ and $h$ such that
$$
1=\sum_jg_jj+h(1-tf),
\qquad j\in J.
$$
Substitute $t=f^{-1}$ in the localization $k[x_1,\ldots,x_n]_f$. This gives $1\in J_f$; clearing a power of $f$ yields
$$
\boxed{\ f^N\in J\text{ for some }N\geq1.\ }
$$
This is the <Strong Hilbert Nullstellensatz> inclusion $I(Z(J))\subseteq\sqrt J$.
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