The following stuff handles a system where no electric charge exists(i.e. free space).
$$\operatorname{rot}\boldsymbol{H}_{}=\sigma\boldsymbol{E}_{}+\epsilon{\partial\boldsymbol{E}_{}\over\partial\mathrm{t}}\tag{1}$$
$$\operatorname{rot}\boldsymbol{E}_{}=-\mu{\partial\boldsymbol{H}_{}\over\partial\mathrm{t}}\tag{2}$$
$$\color{fuchsia}{-\operatorname{rot}\operatorname{rot}\boldsymbol{E}_{}=\sigma\mu{\partial\boldsymbol{E}_{}\over\partial\mathrm{t}}+\epsilon\mu{\partial^2\boldsymbol{E}_{}\over\partial\mathrm{t}^2}}\tag{3}$$
The book says that eqn 3 can be obained differentiating eqn1 by$~t~$and substituting eqn2 to that eqn to eliminate$~\boldsymbol{H}_{}~$
So I started from the following.
$${\partial\over\partial\mathrm{t}}\left(\operatorname{rot}\boldsymbol{H}_{}\right)={\partial\over\partial\mathrm{t}}\left(\sigma\boldsymbol{E}_{}+\epsilon{\partial\boldsymbol{E}_{}\over\partial\mathrm{t}}\right)\tag{4}$$
My brain has freezed to proceed operations from here.
$$\boldsymbol{H}_{}:=\text{magnetic field vector}\tag{5}$$
$~r,\theta~$may determine a value of this vector. And these 2 parameters may be determined by one independent variable$~t~$
Following given list may help to resolve what I want to do.
$$ \begin{cases} \operatorname{rot}\boldsymbol{H}_{}=\boldsymbol{J}_{}+{\partial\boldsymbol{D}_{}\over\partial\mathrm{t}}\\ \operatorname{rot}\boldsymbol{E}_{}=-{\partial\boldsymbol{B}_{}\over\partial\mathrm{t}}\\ \operatorname{div}\boldsymbol{D}_{}=\sigma\\ \operatorname{div}\boldsymbol{B}_{}=0\\ \boldsymbol{D}_{}=\epsilon\boldsymbol{E}_{}\\ \boldsymbol{B}_{}=\mu\boldsymbol{H}_{}\\ \boldsymbol{J}_{}=\sigma\boldsymbol{E}_{}\\ \operatorname{div}\boldsymbol{E}_{}=0\\ \operatorname{div}\boldsymbol{H}_{}=0\\ \epsilon,\mu,\sigma\leftarrow~~\text{constants} \end{cases}\tag{6} $$
I need your advice.
In the first place , how can I differentiate with vector of $~\operatorname{rot}~$?
Which article(s) is/are suitable to be read to resolve my problem?
This is a standard excersise. The key thing to notice is that the curl operator and the time partial derivative can commute because of Clairaut's Theorem, as explained in this post. In your case, this is applied to get $$ \frac{\partial}{\partial t}\left(\nabla \times \vec{H} \right) =\nabla \times \left(\frac{\partial}{\partial t}\vec{H} \right) \tag{0} $$ and thus \begin{align} \frac{\partial}{\partial t}\left(\nabla \times \vec{H} \right) &= \frac{\partial}{\partial t}\left(\sigma \vec{E} + \varepsilon\frac{\partial \vec{E}}{\partial t} \right)\\ \overset{(0)}{\mathbin{\color{blue}{\implies}}}\nabla \times \left(\color{purple}{\frac{\partial}{\partial t}\vec{H}} \right) &=\sigma \frac{\partial \vec{E}}{\partial t} + \varepsilon\frac{\partial^2 \vec{E}}{\partial t^2} \\ \overset{(2)}{\mathbin{\color{blue}{\implies}}}\nabla \times \left(\color{purple}{-\frac{1}{\mu} \nabla \times \vec{E}} \right) &=\sigma \frac{\partial \vec{E}}{\partial t} + \varepsilon\frac{\partial^2 \vec{E}}{\partial t^2} \\ \mathbin{\color{blue}{\implies}}-\nabla \times \left( \nabla \times \vec{E} \right) &=\sigma \mu\frac{\partial \vec{E}}{\partial t} + \varepsilon\mu\frac{\partial^2 \vec{E}}{\partial t^2} \end{align} where you get the desired result, remembering that you can factor out $\varepsilon$ and $\mu$ from the derivatives since they're constants.