Suppose we have two functions $Q=Q(q,p)$ and $p=p(q,Q)$ (the context is not important here, but if you're wondering $(p,q)$ arise as coordinates in a Hamiltonian system, and $(P,Q)$ are alternative coordinates derived from a canonical transformation). I'm looking at an example and the final step is: \begin{equation} \dots=\frac{\partial{Q}}{\partial{p}}\Bigg{|}_q\frac{\partial{p}}{\partial{Q}}\Bigg{|}_q = 1 \end{equation}
Now I have looked at various stackexchange posts and spent quite a while trying understand why you can treat normal derivatives like fractions, and why you can't do the same for partial derivatives, which perhaps seems contradictory to the last step above. I have tried to justify it in the following way, can someone please tell me if this is correct (I've taken out the vertical restriction bars to save time). We have: \begin{equation} dQ=\frac{\partial{Q}}{\partial{q}}dq + \frac{\partial{Q}}{\partial{p}}dp~~~~~,~~~~~dp=\frac{\partial{p}}{\partial{q}}dq + \frac{\partial{p}}{\partial{Q}}dQ \end{equation}
Now, since $q$ is held constant, $dq=0$. Hence if we divide through by $dp$ in the left equation and $dQ$ in the right we get $\frac{\partial{Q}}{\partial{p}}=\frac{dQ}{dp}$ and $\frac{\partial{p}}{\partial{Q}}=\frac{dp}{dQ}$, and using the fact that we can cancel total derivatives like fractions (for reasons I'm not wanting to discuss here, unless they are important to the dicussion), we have:
\begin{equation} \frac{\partial{Q}}{\partial{p}}\frac{\partial{p}}{\partial{Q}} = \frac{dQ}{dp}\frac{dp}{dQ}=1 \end{equation}
Is this correct, and if so, am I overcomplicating things at all? Is there perhaps a more intuitive reason why in this case, we can cancel the partial derivatives like fractions? Thanks