Derivative of Gibbs Energy wrt Temperature at constant Internal Energy I came across with the following solution from Sandler's book: $$\begin{eqnarray}\left( \frac{\partial G}{\partial T} \right)_U &amp;=&amp; \frac{\partial (G,U)}{\partial (T,U)} \\ &amp;=&amp; \frac{\partial (G,U)}{\partial (T,V)}\cdot \frac{\partial (T,V)}{\partial (T,U)} \\ &amp;=&amp;\frac{(\partial G/\partial T)_V (\partial U/\partial V)_T-(\partial G/\partial V)_T (\partial U/\partial T)_V}{(\partial U/\partial V)_T} \end{eqnarray}$$

Please help me understand these steps. What happens in line 1? I know that line 2 is by using Chain Rule by then how did line 2 becomes line 3?
The starting point is the two equations $$dG=\left(\frac{\partial G}{\partial T}\right)_VdT+\left(\frac{\partial G}{\partial V}\right)_TdV$$ and $$dU=\left(\frac{\partial U}{\partial T}\right)_VdT+\left(\frac{\partial U}{\partial V}\right)_TdV$$From the second equation, you determine the partial of V with respect to T and constant U by setting dU=0.  I leave the rest of the math up to you.

Chet