I asked a similar question on phys-stack-ex framed in a thermodynamics context but got no reply. Here I am asking it without the thermo baggage.
Let us consider a total differential with three variables : $$dU=z_0dx_0+z_1dx_1+z_2dx_2 \tag {1} \label{eq1}.$$ Here $x_0, x_1, x_2$ are the independent variables on which the others depend so that: $U = U(x_0,x_1,x_2),..., z_2 = z_2(x_0,x_1,x_2)$. The equality of the mixed partial derivatives $$\begin{align} \frac{\partial ^2 U}{\partial x_0\partial x_1} &=\frac{\partial z_0}{\partial x_1} = \frac{\partial z_1}{x_0}, \\ \frac{\partial ^2 U}{\partial x_2\partial x_1} &=\frac{\partial z_2}{\partial x_1} = \frac{\partial z_1}{x_2}\end{align}, \\etc., \tag {A} \label{eqA}$$ assures the existence of such function $U$.
Rewrite $\eqref{eq1}$ as $$y_0dx_0 = dU -y_1dx_1 - y_2dx_2 \tag {2} \label{eq2}$$ but here the independent variables are taken to be $U, x_1, x_2$ while the others are functions of them such that $x_0=x_0(U,x_1,x_2), ..., y_2=y_2(U,x_1,x_2)$.
Now the RHS of $\eqref{eq2}$ $$\tilde \omega = dU -y_1dx_1 - y_2dx_2 \tag {3} \label{eq3}$$ is a Pfaffian 1st order form of which we know that it has an integrating factor, namely $y_0$. In other words, $\vartheta dU - \vartheta y_1dx_1 -\vartheta y_2dx_2 $ for $\vartheta=\frac{1}{y_0}, y_0\ne 0$ is a total differential $$dx_0 = \vartheta dU - \vartheta y_1dx_1 -\vartheta y_2dx_2 \tag {4} \label{eq4}$$ and that can be the case if and only if we have the following "reciprocity" relationships among the mixed partial derivatives of the $\vartheta, \vartheta y_1, \vartheta y_2$: $$\begin{align} \frac{\partial \vartheta}{\partial x_1} &= - \frac{\partial (\vartheta y_1)}{\partial U}\tag{5}\label{eq5}\\ \frac{\partial \vartheta}{\partial x_2} &= - \frac{\partial (\vartheta y_2)}{\partial U}\tag{6}\label{eq6}\\ \frac{\partial (\vartheta y_1)}{\partial x_2} &= \frac{\partial (\vartheta y_2)}{\partial x_1}\tag{7}\label{eq7} \end{align}.$$ Next we eliminate $\vartheta$ and obtain a differential equation between $y_1$ and $y_2$. Multiply $\eqref{eq5}$ by $y_2$ and $\eqref{eq6}$ by $y_1$, and then add these and the third equation the result of which after cancellations is: $$ \frac{\partial y_1}{\partial x_2} - \frac{\partial y_2}{\partial x_1} = y_1 \frac{\partial y_2}{\partial U} - y_2 \frac{\partial y_1}{\partial U} \tag{8}\label{eq8}$$ (The same result can be had using the Frobenius integration theorem [1] according to which the integrating factor exists iff $\tilde \omega \wedge d\tilde \omega = 0$; this is an easier and more systematic way of generating similar equations for more than three variables and various other potentials resulting from the Legendre transforms.)
It seems to me that $\eqref{eq8}$ is independent of the pair-wise "reciprocity" relations, $\eqref{eqA}$, at least I could not derive it from those.
My question is if it is really true that Equation $\eqref{eq8}$ contains essentially different information from the equality of the pair-wise mixed partials $\eqref{eqA}$ or it can be derived from them? Thank you.
[1]: Flanders: Differential Forms with Applications to the Physical Sciences, section 7.3
If I'm understanding correctly, you're setting $$ z_i(x_0,x_1,x_2) = y_i(U(x_0,x_1,x_2),x_1,x_2), \qquad i =1,2. $$ Then $$ \frac{\partial z_1}{\partial x_2} = \frac{\partial y_1}{\partial U}\frac{\partial U}{\partial x_2} + \frac{\partial y_1}{\partial x_2} = y_2\frac{\partial y_1}{\partial U}+\frac{\partial y_1}{\partial x_2} $$ and likewise $\dfrac{\partial z_2}{\partial x_1} = y_1\dfrac{\partial y_2}{\partial U} + \dfrac{\partial y_2}{\partial x_1}$. From this, it follows that the second of equations (A) $\dfrac{\partial z_1}{\partial x_2} = \dfrac{\partial z_2}{\partial x_1}$ is equivalent to (8).