I have a function $H=f(T,P)$ at which I evaluate $H_1$ at $(T_1,P_1)$. Now I change $P$ to $P_2$, keeping $H$ the same so I can calculate the new value $T_2$. So $f(T_1,P_1)=f(T_2,P_2)$ and I know $T_1,P_1,P_2$ which gives me $T_2$.
Now I have an extra complexity where $T_1$ is a function of another variable, $Q$, so $T_1=g(Q)$. I know $\cfrac{dT_1}{dQ}$, and also $\cfrac{\partial f}{\partial T}$ and $\cfrac{\partial f}{\partial P}$
I need to find $\cfrac{dT_2}{dQ}$.
Set $h(x) = f(x,P_2)$. Then $T_2(Q) = h^{-1}(f(T_1(Q),P_1))$. So, using the chain rule several times, \begin{align} T_2'(Q) &= (h^{-1})'(f(T_1(Q),P_1))\cdot f'(T_1(Q),P_1)\binom{T_1'(Q)}0\\ &= \frac 1{h'(h^{-1}(f(T_1(Q),P_1)))}\cdot\frac{\partial f}{\partial T}(T_1(Q),P_1)T_1'(Q)\\ &= \frac{\frac{\partial f}{\partial T}(T_1(Q),P_1)T_1'(Q)}{\frac{\partial f}{\partial T}(h^{-1}(f(T_1(Q),P_1)),P_2)}\\ &= \cfrac{T_1'(Q).\cfrac{\partial f}{\partial T}(T_1(Q),P_1)}{\cfrac{\partial f}{\partial T}( T_2(Q),P_2)} \end{align}