= Solution
Put $X=[\mathrm O_3]$ and $u=T^{-1}$. The fast <oxygen>–<ozone> partition obeys
$$
k_4[\mathrm O][\mathrm O_2][\mathrm M]=J_5X,
\qquad [\mathrm O]=\frac{J_5X}{k_4[\mathrm O_2][\mathrm M]}.
$$
Dividing the production–loss balance by two and substituting gives
$$
P=AX^2+BX,\qquad
P=J_1[\mathrm O_2],\quad
A=\frac{k_2J_5}{k_4[\mathrm O_2][\mathrm M]},\quad
B=k_3[\mathrm{NO}].
$$
Under the specified restriction, all <temperature> dependence lies in the three rate coefficients: hold the <photodissociation> rates and the <concentrations> of $\mathrm O_2$, $\mathrm M$ and NO fixed. Then $d\ln A/du=-2100-510=-2610\ \mathrm K$ and $d\ln B/du=-1200\ \mathrm K$. Defining $Y=d\ln X/du$, differentiation yields
$$
0=AX^2(-2610+2Y)+BX(-1200+Y).
$$
Therefore
$$
\boxed{\frac{d\ln[\mathrm O_3]}{d(T^{-1})}
=\frac{2610L_C+1200L_N}{2L_C+L_N},\qquad
L_C=k_2[\mathrm O][\mathrm O_3],\quad L_N=k_3[\mathrm{NO}][\mathrm O_3].}
$$
The factor two in the denominator multiplying $L_C$ comes from the additional dependence $[\mathrm O]\propto X/k_4$; ignoring that dependence would give the wrong sensitivity. This derivative has units of kelvin.
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