Showing that $x\cdot w_x + y\cdot w_y = z \cdot w_z, w = F(xz, yz)$ We'd like to show that $x\cdot w_x + y\cdot w_y = z \cdot w_z, w = F(xz, yz)$, where $w_x$ is the partial derivative of $w$ w.r.t. x (resp. for $y$ and $z$).
It seems that I'm bit rusty with my multivariable calculus. Given that $w = F(xz, yz)$, I'm not really sure how to "take" the partial derivatives of $F$ w.r.t. the components $x, y, z$, as the form of $F$ is not specified in the problem. Namely, if we let $f: \mathbb{R}^3 \to \mathbb{R}, (x, y, z) \mapsto (xz, yz)$, then  we'd like to first determine the partial derivatives of $(F \circ f)$ w.r.t. $x, y$ and $z$. We can quite easily find the Jacobian of $f$, as it is at a point $a \in \mathbb{R}^3$ equal to
$$f'(a) = \begin{bmatrix}\frac{\partial xz}{\partial x} & \frac{\partial xz}{\partial y} & \frac{\partial xz}{\partial z}\\
\frac{\partial yz}{\partial x} & \frac{\partial yz}{\partial y} & \frac{\partial yz}{\partial z}\end{bmatrix} = \begin{bmatrix}z & 0 & x\\ x & z & y\end{bmatrix}$$
But I'm not really sure how to proceed beyond this point.
Edit:
 A: Define$$\begin{array}{rccc}\psi\colon&\Bbb R^3&\longrightarrow&\Bbb R^2\\&(x,y,z)&\mapsto&(xz,yz).\end{array}$$Then $\psi$ is differentiable and, if $(x,y,z)\in\Bbb R^3$,$$\psi'(x,y,z)(\alpha,\beta,\gamma)=(z\alpha+x\gamma,z\beta+y\gamma).$$So,
$$w_z(\alpha,\beta,\gamma)=\alpha\frac{\partial F}{\partial x}(x\gamma+z\alpha)+\beta\frac{\partial F}{\partial y}(z\beta+y\gamma).$$We also have$$w_x(\alpha,\beta,\gamma)=\gamma\frac{\partial F}{\partial x}(x\gamma+z\alpha)\quad\text{and}\quad w_y=\gamma\frac{\partial F}{\partial y}(z\beta+y\gamma)$$and therefore$$\gamma w_z(\alpha,\beta,\gamma)=\alpha w_x(\alpha,\beta,\gamma)+\beta w_y(\alpha,\beta,\gamma)$$indeed.
A: A very expressive way:
$$
x\,w_x + y\,w_y - z\,w_z = 0,
$$
can be written in vectorial form as
$$
\begin{pmatrix}x \\ y \\ -z\end{pmatrix} \cdot \begin{pmatrix}w_x \\ w_y \\ w_z\end{pmatrix} = 0
$$
Hence, $(x, y, -z)^T$ is orthogonal to $\nabla w$. Let $w$ be parametrized by
$$
M(\xi, \eta, \zeta) = \begin{pmatrix}x(\xi, \eta, \zeta)\\ y(\xi, \eta, \zeta)\\ z(\xi, \eta, \zeta)\\ w(\xi, \eta, \zeta)\end{pmatrix},
$$
then we can build a vector on the tangent plane such that
$$
\frac{d x}{d \zeta} = x, \qquad x(\xi, \eta, 0) = x_0(\xi, \eta),
$$
$$
\frac{d y}{d \zeta} = y, \qquad y(\xi, \eta, 0) = y_0(\xi, \eta),
$$
$$
\frac{d z}{d \zeta} = -z, \qquad z(\xi, \eta, 0) = z_0(\xi, \eta).
$$
$$
\frac{d w}{d \zeta} = 0, \qquad w(\xi, \eta, 0) = w_0(\xi, \eta).
$$
Solving the system,
$$
x(\xi, \eta, \zeta) = x_0(\xi, \eta) e^\zeta,
$$
$$
y(\xi, \eta) = y_0(\xi, \eta) e^\zeta, 
$$
$$
z(\xi, \eta) = z_0(\xi, \eta) e^{-\zeta},
$$
$$
w(\xi, \eta, \zeta) = F(\xi, \eta).
$$
If we can invert $(\xi, \eta, \zeta) \to (x, y, z)$, we will have found the general solution to the PDE that passes trough $(x_0, y_0, z_0, w_0)$.
For the special case $x_0(\xi, \eta) = \xi$, $y_0(\xi, \eta) = \eta$, $z_0(\xi, \eta) = 1$, $w_0(\xi, \eta) = F(\xi, \eta)$,
$$
\xi = x z, \quad \eta = y z, \quad \zeta = -\ln z
$$
and then
$$
w(x, y, z) = F(x z, y z).
$$
Note that we have solved the following initial value problem:
$$
x\,w_x + y\,w_y - z\,w_z = 0, \qquad w(x, y, 1) = F(x, y)
$$
