Calculation of second order partial derivatives Let $$F\left( \frac{x}{z}, \frac{y}{z} \right)=0$$ determine a function $z=z(x,y)$. Show that $$\frac{\partial^2z}{\partial x^2} \cdot \frac{\partial^2z}{\partial y^2}-\left( \frac{\partial^2z}{\partial x\partial y} \right)^2=0.$$
I was frustrated in simplifying doing the second partial derivatives...
 A: In order for the definition of $z(x,y)$ to make sense, we must assume that on each line through $(0,0)$ there is exactly one point with $F(x,y)=0$. This is possible (for example the locus of $F(x,y)$ could be a parabola with the origin on the inside), but in order for $z$ to be continuous everywhere, it seems to be more meaningful to suppose that each ray emanating from $(0,0)$ contains exactly one zero of $F$, and require that $z$ is positive.
Now let's first prove the property for the half-plane $x>0$:
Define $f(m)$ to be the reciprocal of the $x$-coordinate of the point on the line $y=mx$ (where $x>0$) that makes $F$ vanish. We can then write $z$ as
$$ z(x,y) = x f(y/x) $$
Assuming that $f$ is sufficiently smooth (otherwise we're in probably insurmountable trouble anyway), it is now easy to write down the partial derivatives of $z$ in terms of $x$, $y$, and the first few derivatives of $f$. After several terms cancel, I get
$$ \frac{\partial^2 z}{\partial x^2} = \frac{y^2}{x^3}f''(y/x)
\qquad \frac{\partial^2 z}{\partial x\partial y} = \frac{-y}{x^2}f''(y/x)
\qquad \frac{\partial^2 z}{\partial y^2} = \frac1x f''(y/x) $$
and it is now trivial to verify that the claimed identity holds.
This argument can be reused identically for the half-plane $x<0$ (except that $f$ will be different), and will again work for $y>0$ and $y<0$, by swapping every $x$ for $y$ and vice versa (which fortunately leaves the target identity unchanged).
This handles everything except for $(x,y)=(0,0)$, but $z$ can't really be defined there anyway, or at least not continuously unless $F=0$ in a neighborhood of $(0,0)$, so let's just assume we're not meant to handle this point anyway.
