The subtility lies in the fact that $\tau(\epsilon)$ and the integral is on k. That means you can transform $$ \left(\frac{\partial f}{\partial \epsilon} \right) = \delta(\epsilon - \epsilon_F) $$ With it, you know you'll only evaluate $\tau(\epsilon)=\tau(\epsilon_F)$ so you change it. But since you integrate on k instead of $\epsilon$, the integral doesn't disappear. Then, you use the identity (13.26) (stated in the question) to solve the problem.

The subtility lies in the fact that $\tau(\epsilon)$ and the integral is on k. That means you can transform $$ \left(\frac{\partial f}{\partial \epsilon} \right) = \delta(\epsilon - \epsilon_F) $$ With it, you know you'll only evaluate $\tau(\epsilon)=\tau(\epsilon_F)$ so you change it. But since you integrate on k instead of $\epsilon$, the integral doesn't disappear. You can retransform the $\delta(\epsilon - \epsilon_F)$ into the Fermi function using the same identity. With the Fermi function, you use the identity (13.26) (stated in the question) to solve the problem. There's a subtility right here $$ \sigma = e² \tau(\epsilon_F) \int \frac{dk}{4 \pi³} \left(-\frac{\partial } {\partial k} f(\epsilon (k)) \right)v(k) $$ where you need to use $$ \int_C dr (u\nabla v) = -\int_C dr (v\nabla u) \tag{I.1} $$ which is true if u(r) and v(r) have the periodicity of the Bravais lattice and C is taken over a primitive cell C.