Equal partial derivatives Suppose $\frac{\partial}{\partial x}F(x,y,z)=\frac{\partial}{\partial y}F(x,y,z)=\frac{\partial}{\partial z}F(x,y,z)$. Does this imply $F(x,y,z)=G(x+y+z)$ for some function G?
 A: The statement is true if all 3 partial derivatives exist and continuous over all 
$\mathbb{R}^3$. 
The proof relies on a multivariable differentiability theorem:

Let $\varphi : U \to \mathbb{R}\;$ be a function defined on some open
  subset $U$ of $\mathbb{R}^m$. If all $m$ partial derivatives of
  $\varphi$ exists and continuous over $U$, then $\varphi$ is
  differentiable over $U$.

If all 3 partial derivatives $\frac{\partial F}{\partial x}$,
$\frac{\partial F}{\partial y}$, $\frac{\partial F}{\partial z}$ exists and continuous
over some open convex set K, then by above theorem, $F$ is differentiable over $K$.
Now for any two point $\vec{v}_1 = (x_1,y_1,z_1),\;\;\vec{v}_2 = (x_2, y_2, z_2) \in K$ such that $$x_1+y_1+z_1 = x_2 + y_2+z_2,$$
the line segment between $\vec{v}_1$ and $\vec{v}_2$:
$$[0,1] \in t\quad\mapsto\quad\vec{v}(t) = \vec{v}_1 + (\vec{v}_2 - \vec{v}_1) t$$
lies completely in $K$. Consider following two functions defined on $[0,1]$:
$$\begin{cases}
f(t) = F(\vec{v}(t))\\
\\
g(t) = 
\frac{\partial F}{\partial x}(\vec{v}(t)) =
\frac{\partial F}{\partial y}(\vec{v}(t)) =
\frac{\partial F}{\partial z}(\vec{v}(t))
\end{cases}
$$
Since $F$ is differentiable over $K$, we can express the derivative of $f$ in terms of
partial derivatives of $F$ and hence in $g$:
$$\begin{align}\frac{df(t)}{dt} = &
  (x_2-x_1) \frac{\partial F}{\partial x}(\vec{v}(t))
+ (y_2-y_1) \frac{\partial F}{\partial y}(\vec{v}(t))
+ (z_2-z_1) \frac{\partial F}{\partial z}(\vec{v}(t))\\
= & ((x_2+y_2+z_2) - (x_1+y_1+z_1)) g(t)\\
= & 0
\end{align}$$
From this, we can conclude
$$f(0) = f(1) \quad\iff\quad F(\vec{v}_1) = F(\vec{v}_2)$$
i.e. $F(\vec{v})$ is constant on the intersection of $K$ with the planes $x+y+z = \text{constant}$. 
If $K$ is the whole $\mathbb{R}^3$, then
$$F(x,y,z) = F(x+y+z,0,0)$$
and the statement follows immediately.
A: Partial Answer:
I tend to think that it is true, but not sure. For polynomials it is true:
Prove by induction on the grade $n$.
For $n=0$ it is true obiously.
Suppose $P_n$ is a polynomial of grade $n$ and has the property:
$$
\partial_xP_n=\partial_yP_n=\partial_zP_n=Q_{n-1}
$$ 
The vector field $(Q_{n-1},Q_{n-1},Q_{n-1})$ is a gradient and then its rotor is null. This implies 
$$
\partial_xQ_{n-1}=\partial_yQ_{n-1}=\partial_zQ_{n-1}
$$
which implies by induction that $Q_{n-1}=q(x+y+z)$.  $\;\square$

Update:
In dimmension 2 it is true (and I suppose in more dimmensions also). Suppose $f$ is such that $f_x=f_y$. Then $f$ is the solution of the following differential problem:
$$
u_x(x,y)-u_y(x,y)=0 \\ u(x,0)=f(x,0)=g(x)
$$
We can solve this problem with the method of the caractheristic. We obtain that $u(x,y)=g(x+y)$. 

Update 2:
Fix z. Then using the previous result we conclude: $f(x,y,z)=g(x+y,z)=g(u,z)$. Now fix $y=0$. Then $f_x=f_z$ implies that $g_u=g_z$ and we conclude that $g(u,z)=h(u+z)$ and we are done.
