Reconstruct a function $f: \mathbb{R}^3 \to \mathbb{R}$ from two identities on its partial derivatives I'd like to find some non-constant function $f: \mathbb{R}^3 \to \mathbb{R}$ such that
$$
\begin{cases}
    \displaystyle \frac{ \partial f } { \partial y } = -2 \frac{ \partial f } { \partial x }, \\[4pt]
    \displaystyle \frac{ \partial f } { \partial z } = - \frac x z \frac{ \partial f } { \partial x }.
\end{cases}
$$
If it helps, it can be assumed that $f$ is as smooth as needed. 
Given the nature of the underlying physical problem, I initially tried looking for solutions of the form $f(x,y,z) = ( a x + b y ) z^c$, to no avail (I get $f=0$). I then tried the more general form $f(x,y,z) = g(x,y) h(z)$, which did not pan out either. I tried more complicated forms using some symbolic computation software, but I could not find a suitable non-constant answer.
I am starting to suspect that there are no non-constant functions satisfying these identities. However, I do not see how to go about proving that.
My question is: can we find a non-constant function satisfying these identities? If not, how can we prove that the only solution would be a constant function?
 A: I am afraid constant functions are the only possible solutions to your system.
Your system reads equivalently
\begin{align}
2\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}&=0,\\
x\frac{\partial f}{\partial x}+z\frac{\partial f}{\partial z}&=0.
\end{align}
The first equation implies that
$$
\frac{\rm d}{{\rm d}t}f(2t+x_0,t,z_0)=2\frac{\partial f}{\partial x}(2t+x_0,t,z_0)+\frac{\partial f}{\partial y}(2t+x_0,t,z_0)=0
$$
holds for all $x_0$ and $z_0$. Thus
$$
f(2t+x_0,t,z_0)=f(x_0,0,z_0).
$$
Define
$$
g(u,v)=f(u,0,v).
$$
Then
$$
f(2t+x_0,t,z_0)=g(x_0,z_0).
$$
Perform the change of variables $x=2t+x_0$, $y=t$ and $z=z_0$, and the last formula yields
$$
f(x,y,z)=g(x-2y,z).
$$
So far, this is a necessary condition for $f$, i.e., if $f$ satisfies the first equation in your system, it must be able to be expressed by some bivariate function $g$ as above. Conversely, you may also check that for any differentiable function $g$, $f(x,y,z)=g(x-2y,z)$ as a trivariate function always satisfies the first equation in your system. Therefore, we conclude that

A trivariate function $f=f(x,y,z)$ satisfies the first equation in your system, if and only if there exists some bivariate differentiable function $g=g(u,v)$, such that
  $$
f(x,y,z)=g(x-2y,z).
$$

Thanks to the above conclusion, we are able to deal with the second equation in your system. Note that
\begin{align}
\frac{\partial f}{\partial x}(x,y,z)&=\frac{\partial}{\partial x}g(x-2y,z)=\frac{\partial g}{\partial u}(x-2y,z),\\
\frac{\partial f}{\partial z}(x,y,z)&=\frac{\partial}{\partial z}g(x-2y,z)=\frac{\partial g}{\partial v}(x-2y,z).
\end{align}
Thus the second equation in your system reads equivalently
$$
x\frac{\partial g}{\partial u}(x-2y,z)+z\frac{\partial g}{\partial v}(x-2y,z)=0.
$$
Perform another change of variables $u=x-2y$ and $v=z$, and the above equation reads
$$
\left(u+2y\right)\frac{\partial g}{\partial u}(u,v)+v\frac{\partial g}{\partial v}(u,v)=0.
$$
This last equation must hold for all $u$, $v$ and $y$ (since $x$, $y$ and $z$ are completely arbitrary, so are $u$ and $v$). However, the arbitrariness of $y$ forces
$$
\frac{\partial g}{\partial u}(u,v)=0,
$$
which in turn, as per the arbitrariness of $v$, forces
$$
\frac{\partial g}{\partial v}(u,v)=0.
$$
These necessary conditions require
$$
g\equiv\text{const}.
$$
In other words, if $f(x,y,z)=g(x-2y,z)$ satisfies the second equation in your system, it is a must that $g$ is constant. Conversely, a constant function obviously satisfies the second equation in your system. Thus we obtain

A trivariate function $f=g(x-2y,z)$ satisfies the second equation in your system, if and only if $g$ is a constant function.

Finally, combined with the two conclusions from above, we eventually have

A trivariate function $f=f(x,y,z)$ satisfies both equations in your system, if and only if $f$ is a constant function.

