Given z_(xy)/(z_xz_y)=φ(z) , show u=z_x/z_y satisfies…

Given $z_{xy}/(z_xz_y)=\phi(z)$, show $u=z_x/z_y$ satisfies $uu_{xy}=u_xu_y$.

Step 1 · Reformulate the target identity

Since $(\ln u)_{xy}=u_{xy}/u-u_xu_y/u^2$, the identity $uu_{xy}=u_xu_y$ is equivalent to $(\ln u)_{xy}=0$.

Step 2 · Expand $(\ln u)_{xy}$

With $\ln u=\ln z_x-\ln z_y$: \[ (\ln u)_{xy} = \frac{z_{xxy}}{z_x}-\frac{z_{xx}z_{xy}}{z_x^2} -\frac{z_{xyy}}{z_y}+\frac{z_{xy}z_{yy}}{z_y^2}. \]

Step 3 · Differentiate the hypothesis and substitute

Differentiating $z_{xy}=\phi(z)z_xz_y$ in $x$ and in $y$ gives \[ \frac{z_{xxy}}{z_x} = \phi'z_xz_y+\phi\frac{z_{xx}}{z_x}z_y+\phi z_{xy}, \qquad \frac{z_{xyy}}{z_y} = \phi'z_xz_y+\phi z_{xy}+\phi\frac{z_{yy}}{z_y}z_x. \] Substituting both into the expansion above, every term cancels in pairs, leaving $(\ln u)_{xy}=0$.

Verification on examples

For $z=xy$: $z_{xy}/(z_xz_y)=1/(xy)=1/z$ (a function of $z$); $u=y/x$ gives $uu_{xy}=(y/x)(-1/x^2)=-y/x^3$ and $u_xu_y=(-y/x^2)(1/x)=-y/x^3$. \checkmark For $z=x^2y$: $z_{xy}/(z_xz_y)=1/(x^2y)=1/z$; $u=2y/x$ gives $uu_{xy}=u_xu_y=-4y/x^3$ in both cases. \checkmark