Derivation for Specific heat of gas at Constant Volume The mathematical interpretation for first law of thermodyanamics is
$$dU = \delta{Q} - \delta{W}$$
for quasi-static process,
$$W = PdV$$
so,
$$dU = \delta{Q} - PdV$$
Constant volume $dv = 0$
so,
$$dU = \delta{Q}$$
we also know that $U = U(T,v)$
using chain rule we get,
$$dU = \frac{\partial{U}}{\partial{T}}\cdot dT + \frac{\partial{U}}{\partial{V}}\cdot dV$$
again for constant volume, $dv = 0$
$$dU = \frac{\partial{U}}{\partial{T}}\cdot dT$$
so from above equations,
$$\frac{\partial{U}}{\partial{T}}\cdot dT = \delta{Q}$$
$$\implies \frac{\partial{U}}{\partial{T}} = \frac{\partial{Q}}{\partial{T}}$$
Now here comes my question, how did the conversion from $\delta{Q} $ and $dT$ to $\frac{\partial{Q}}{\partial{T}}$ take place?
 A: As $dU=\delta Q$
$$\implies \frac{dU}{dT}=\frac{\delta Q}{\delta T}$$
As $U(V,T)$
So, $$\frac{dU}{dT}=\frac{\partial U}{\partial T}|_V\frac{dT}{dT}+\frac{\partial U}{\partial V}|_T\frac{dV}{dT}$$
For constant volume, $\frac{dV}{dT}=0$
So, $$\frac{dU}{dT}=\frac{\partial U}{\partial T}|_V\frac{dT}{dT}=\frac{\partial U}{\partial T}|_V$$
So, $$\frac{\partial U}{\partial T}|_V=\frac{\delta Q}{\delta T}$$
There should not be $\frac{\partial Q}{\partial T}$ in the RHS of final expression.
A: Your question is about concepts and formalism, in particular, usage of $\delta$ for the so-called inexact differentials, may be confusing.
Let me restate the problem without using $\delta$, but using the same notation used for example by Planck, who knew quite well Thermodynamics.
The first principle of Thermodynamics says that it is possible to define a function of state $U$, such that
$$
\Delta U_{AB} = w + q
$$
where $w$ and $q$ are the work and heat exchanged in the process driving the passage from the equilibrium state $A$ to another equilibrium state $B$. I am using the physical sign convention where $w$ is positive if it increases the energy of the system. Just a change of sign allows to recover the convention used in the question. $U$ is a function of the state. Therefore, for every two states $A$ and $B$ close enough, it is possible to approximate $\Delta U$ by its differential within an error which is second order in the differences between the independent state variables (this is a math theorem valid for every differentiable function with continuous first derivatives).
Thus, for close states $A$ and $B$, and using $T$ and $V$ and state variables, for a constant volume transformation we have
$$
\Delta U = \left. \frac{\partial U}{\partial T} \right|_V \Delta T + 
\mathcal{O}(\Delta T^2).
$$
Now, for a simple system, macroscopic work on the system is zero if volume is constant. Therefore we also have
$$
\Delta U = q.
$$
This is the key point. Therefore, for an isochoric process, heat becomes a function of state exactly like, for an adiabatic process work becomes a function of state.
The (differential) specific heat is proportional to the (extensive) heat capacity defined as the limit for $\Delta T \rightarrow 0$ of the ratio between $q$ and $\Delta T$. Then, in the present case we can write
$$
C_V= \lim_{\Delta T \rightarrow 0} \frac{q(\Delta T)}{\Delta T} = 
\lim_{\Delta T \rightarrow 0} \frac{\Delta U}{\Delta T}=
\left. \frac{\partial U}{\partial T} \right|_V 
$$
