You hold all variables other than $T$ fixed, and then differentiate with respect to $T$.
To see that taking the first partial is easy, note that your expression has the form
$$
p=a_1 T +a_2 T+a_3+a_4T^{-2}+a_5 T+a_6+a_7 T^{-2}
$$
where
$\qquad a_1={R\over V}$,
$\qquad a_2={B_0R\over V^2}$,
$\qquad a_3={-A_0\over V^2}$,
$\qquad a_4={-C_0\over V^2}$,
$\qquad a_5= {bR\over V^3}$,
$\qquad a_6={-a\over V^3}+{A\alpha\over V^6}$,
and
$\qquad a_7={c(1+{\gamma\over V^2}})\cdot{1\over V^3}\cdot e^{-\gamma/V^2}$.
Note that none of the $a_i$ depend on $T$; thus, when taking the first partial of $p$, we are differentiating a sum whose terms are multiples of powers of $T$. We have
$$\tag{1}\eqalign{
{\partial p\over\partial T}
&=a_1+a_2+0-2a_4T^{-3}+a_5+0-2a_7T^{-3}\cr
&=a_1+a_2+a_5-2a_4T^{-3}-2a_7T^{-3}
}
$$
Now take the partial of $(1)$ with respect to $T$ above to obtain the second partial of $p$:
$$
\eqalign{
{\partial^2 p\over\partial T^2}
&={\partial\over\partial T}\bigl(a_1+a_2+a_5-2a_4T^{-3}-2a_7T^{-3}\bigr)\cr
&=6a_4T^{-4}+6a_7T^{-4}\phantom{T\over T}\cr
&=6\cdot{-C_0\over V^2}\cdot{1\over T^4}+6\cdot{c(1+{\gamma\over V^2})}\cdot{1\over V^3}\cdot e^{-\gamma/V^2}\cdot{1\over T^4}\cr
&={6\over V^2T^4}\Bigl( {c\over V}(1+{\gamma\over V^2}) e^{-\gamma/V^2} -C_0\Bigr).
}
$$