Internal Energy Definition, Why is dU zero when dT is zero? According to my Equation sheet, $$\delta U=\left(\frac{\delta U}{\delta T}\right)_V\,\delta T+\left(\frac{\delta U}{\delta V}\right)_T\,\delta V=C_V\,\delta T-C_V\left(\frac{\delta T}{\delta V}\right)_U\,\delta V$$ Where $\delta U$ is change in internal energy. 

So one of the homework solutions said that for a system with a constant Temperature, $\delta T=0$ $\Delta U=0$. And I remember one of the TAs saying that $\Delta U=C_V\,\delta T$ and since $\delta T=0$ the value goes to zero, which makes sense. But she didn't even include the second term: $-C_V\left(\frac{\delta T}{\delta V}\right)_U\,\delta V$. 

I was trying to make sense of the derivative but I'm really having trouble with the second term. The small letter $U$ outside the parenthesis is supposed to mean constant $U$.

Why am I allowed to ignore the second term?

The equation for enthalpy is very similar, and I have the same problem:

$$\delta H=\left(\frac{\delta H}{\delta T}\right)_p\,\delta T+\left(\frac {\delta H}{\delta H}\right)_T\,\delta H=C_p\,\delta T-C_p\left(\frac{\delta T}{\delta p}\right)_H\,\delta p$$
The second term in each of your two equations can be expressed solely in terms of the equation of state for the solid, liquid, or gas:$$\left(\frac{\partial U}{\partial V}\right)_T=-\left[P-T\left(\frac{\partial P}{\partial T}\right)_V\right]$$$$\left(\frac{\partial H}{\partial P}\right)_T=\left[V-T\left(\frac{\partial V}{\partial T}\right)_P\right]$$
In the case of an ideal gas, the right hand sides of both these equations are equal to zero.