Please help me to derive that #((delS)/(delT))_P = C_P/T# and #((delS)/(delT))_V = C_V/T#?

Okay, so I will assume you know or can figure out the Maxwell relations. Those are a common starting point. Furthermore, you have by definition that #C_P = ((delH)/(delT))_P# and #C_V = ((delU)/(delT))_V#, so you do not have to derive those.

Recall that #G = H - TS#. Then, the derivative is:

#dG = dH - d(TS)#

#= dH - SdT - TdS#
(use product rule)

Now, for the first derivation, you can divide through by #delT# at a constant #P# to generate two derivatives that can be figured out, and the third one is what you're looking for:

#((delG)/(delT))_P = ((delH)/(delT))_P - cancel(S((delT)/(delT))_P)^(1) - T((delS)/(delT))_P#

#" "" "bb((1))#

where the third term goes to #1# because #T# does not change with respect to #T#; it is one-to-one with itself.

By definition, #((delH)/(delT))_P = C_P#. #" "" "bb((2))#

Next, recall that for the natural variables #T# and #P# correspond to the Gibbs' free energy, so that the Maxwell relation is:

#dG = -SdT + VdP# #" "" "" "bb((3))#

We do not yet know what #((delG)/(delT))_P# is, but using the Maxwell relation, we get:

#((delG)/(delT))_P = -S# #" "" "" "bb((4))#

Therefore. plugging #bb((4))# and #bb((2))# into #bb((1))#:

#cancel(-S) = C_P - cancel(S) - T((delS)/(delT))_P#

It follows that if #C_P = T((delS)/(delT))_P#, it must be that #color(blue)(((delS)/(delT))_P = C_P/T)#.

A similar process follows for deriving #C_V/T = ((delS)/(delT))_V#. Using:

#A = U - TS#,

take the derivative to get:

#dA = dU - SdT - TdS#

Almost like before, divide by #delT# at a constant #V# instead of #P# to generate two derivatives that can be figured out, and the third one is what you're looking for:

#((delA)/(delT))_V = ((delU)/(delT))_V - Scancel(((delT)/(delT))_V)^(1) - T((delS)/(delT))_V#

We know by definition that #((delU)/(delT))_V = C_V#, #((delT)/(delT))_V = 1#, and that #A# is a function of the natural variables #T# and #V#.

So from the Maxwell relation:

#dA = -SdT - PdV#

we have that:

#((delA)/(delT))_V = -S#,

so plugging back into the main equation, we get:

#cancel(-S) = C_V - cancel(S) - T((delS)/(delT))_V#

Since #C_V = T((delS)/(delT))_V#, it follows that #color(blue)(((delS)/(delT))_V = C_V/T)#.