$S(E,V)$ for an ideal gas So, $$dS = \frac{dE}{T}+\frac{P}{T}$$
For an ideal gas, the above relation reduces to,
$$dS = \frac{C_VdE}{E}+\frac{nR}{V}$$
Integrating both sides, I can obtain,
$$S(E,V) = C_V\ \ln{E}+nR\ \ln{V}+\textrm{constant}$$
That's fine for me, but I am struggling to find same relation from the partial derivative of $S(E,V)$. Using, $$\left ( \frac{\partial S }{\partial V} \right )_E = \frac{P}{T}= \frac{nR}{V}$$
Integrating both sides, I can obtain: $$S(E,V) = nR \ln{V}+f(E)$$
And using,$$\left ( \frac{\partial S }{\partial E} \right )_V = \frac{1}{T}= \frac{C_V}{E}$$ 
Again, integrating, I can obtain:
$$S(E,V) = C_V \ln{E}+f(V)$$
So, comparing these two relations I can say:
$$S(E,V) = nR \ln{V} + C_V\ln{E} $$
So, there is an extra constant term missing, which I can't produce with these two relations. I think that my conundrum is more due to Maths than the underlying physics in here. Would be very grateful if anyone can suggest, what's wrong I am doing in here. Thanks.
 A: You have neglected constants of integration when comparing terms.
You have :
$$S(E,V)=nR\ln V +f(E)$$
and
$$S(E,V)=C_V\ln E +g(V)$$
You assume :
$$f(E)=C_V\ln E$$
but should say :
$$f(E)=C_V\ln E + K_1$$
for some constant $K_1$ and similarly for $g(V)$.
Doing that you get :
$$S(E,V)=nR\ln V + C_V\ln E + K_1$$
And hence your desired result.
A: You can look at it like this: when evaluating $\left ( \frac{\partial S }{\partial E} \right )_V = \frac{C_V}{E}$ you should keep in mind what you already calculated:
$$ \left ( \frac{\partial S }{\partial E} \right )_V = \left ( \frac{\partial\ (nR \ln{V}+f(E))}{\partial E} \right )_V = \left ( \frac{\partial\ nR \ln{V}}{\partial E} \right )_V + \left ( \frac{\partial f(E)}{\partial E} \right )_V = \left ( \frac{\partial f(E)}{\partial E} \right )_V = \frac{C_V}{E}$$
Because V is constant in the first partial derivative, the derivative is $0$. So after integrating you have information about $f(E)$ more so than about $S$. By integration you get:
$$ f(E) = C_V \ln{E}+ \text{constant}$$
This constant can no longer depend on $V$ anymore because you defined $f$ to be only dependant on $E$. Filling in your value for $f$ in the equation you got earlier gives:
$$ S(E,V) = nR \ln{V}+f(E) = nR \ln{V} +  C_V \ln{E}+ \text{constant}$$
