Thermodynamic relation $$\left(\frac{dU}{dT}\right)_p= \left( \frac{\partial U}{\partial T} \right)_v + \left( \frac{\partial U}{\partial V}\right)_T \left( \frac{\partial V}{\partial T}\right)_p$$
Using maxwell relations etc, can you show me how to prove above thermodynamic relation.
 A: When dealing with thermodynamic equalities, it is very useful to introduce Jacobian determinants
$$ \frac{\partial (u,v)}{\partial (x,y)} = \begin{vmatrix} (\partial_x u)_y & (\partial_y u)_x \\
(\partial_x v)_y & (\partial_y v)_x
\end{vmatrix}.$$
Here, we have and will implicitly assume that $u(x,y)$ and $v(x,y)$. But because we often change the variables which the function depends on, we always keep the variable which is kept constant as a subscript to the bracket surrounding the partial derivative. In thermodynamics, you should always think that functions are defined implicitly. All quantities live on a 2D surface. Thus specifying two independent coordinates you can figure out a third one.
The Jacobian have the following relevant properties:
$$\frac{\partial (u,y)}{\partial (x,y)} = \left( \frac{\partial u}{\partial x} \right)_y,$$
$$\frac{\partial (u,v)}{\partial (x,y)} = \left(\frac{\partial (x,y)}{\partial (u,v)} \right)^{-1}, \qquad\text{(inverse function theorem)}$$
and
$$\frac{\partial (u,v)}{\partial (x,y)} = \frac{\partial (u,v)}{\partial (s,t)} \frac{\partial (s,t)}{\partial (x,y)}   . \qquad\text{(chain rule)}$$
In your case
$$\left(\frac{\partial U}{\partial T} \right)_p= \frac{\partial (U,p)}{\partial (T,p)} = \frac{\partial(U,p)/\partial(T,V) }{\partial(T,p)/\partial(T,V)} 
= \frac{(\partial U/\partial T)_V (\partial p/\partial V)_T -
(\partial U/\partial V)_T (\partial p/\partial T)_V}{\left(\partial p/\partial V\right)_T}. $$
Now, we evaluate
$$\frac{(\partial p/\partial T)_V}{(\partial p/\partial V)_T}
= \frac{\partial(p,V)}{\partial(T,V)} \Big/\frac{\partial(p,T)}{\partial(V,T)}
= - \frac{\partial(p,V)}{\partial(V,T)} \frac{\partial(V,T)}{\partial(p,T)}
= - \frac{\partial(p,V)}{\partial(p,T)}  = -\left(\frac{\partial V}{\partial T}\right)_p .$$
So in total, we have
$$\left(\frac{\partial U}{\partial T} \right)_p =\left(\frac{\partial U}{\partial T}\right)_V +\left( \frac{\partial U}{\partial V}\right)_T \left(\frac{\partial V}{\partial T}\right)_p. $$
Note that we did not use any Maxwell relation. Your relation is just a fact about partial derivatives.
A: The quantity $U$ can be considered as a function of $p$ and $T$, let's write it as $U = f(p,T)$, or as a function of $V$ and $T$, let's write that as $U=g(V,T)$. We can also view $V$ as a function of $p$ and $T$, which we write as $V=h(p,T)$, and this gives us the relation between the two ways of representing $U$:
$$
f(P,T)=g(h(p,T),T).
$$
Taking the partial derivative w.r.t. $T$ on both sides, gives (according to the chain rule)
$$
\frac{\partial f}{\partial T}(P,T) =
\frac{\partial g}{\partial V}(h(p,T),T) \cdot \frac{\partial h}{\partial T}(p,T)
+ \frac{\partial g}{\partial T}(h(p,T),T) \cdot 1
,
$$
and this is just another way of writing your relation. (In a pure mathematician's notation, one might say, since (unlike phycisists and applied mathematicians) we avoid using the same letter for a physical quantity and for the mathematical function which represents the dependence of that quantity on other physical quantities. In this case, I would say that this practice reduces the risk of confusion, at least for beginners.)
