Functions where the sum of its partial derivatives is zero I am currently studying Calculus of Variations and have come up with this problem.

What functions $f:\Bbb R^n\to\Bbb R$ satisfy $$\sum\limits_i\frac{\partial f}{\partial x_i}=0\tag1$$ with $f$ of differentiability class $C^\infty$?

It can be shown that if the total derivative is zero; that is, if $$\frac{dF}{dt}=\sum_i\frac{\partial f}{\partial x_i}\cdot\frac{dx_i}{dt}=0$$ with $F(t)=f(x_1(t),\cdots,x_n(t))$, then $f$ is constant. However, I cannot see how the same approach can be used for $(1)$.
The trivial function $f(x)=c$ satisfies $(1)$. For even $n>1$, a non-trivial solution is the function $$f(x_1,\cdots,x_n)=\exp\left(\sum_i(-1)^{i+1}x_i\right)$$ since consecutive partial derivatives of $f$ cancel each other out. I suspect there are more solutions that invoke trigonometric expressions.
Is it possible to derive the entire family/families of functions $f$ that satisfy $(1)$? 
 A: Define $$y_i:=x_i-x_n$$ for every $i=1,2,\ldots,n-1$, and $$y_n:=x_n\,.$$   Then, we have
$$x_i=y_{i}+y_n$$
for each $i=1,2,\ldots,n-1$, and $$x_n=y_n\,.$$  Observe that
$$\frac{\partial}{\partial y_n}=\sum_{k=1}^n\,\left(\frac{\partial x_k}{\partial y_n}\right)\,\frac{\partial}{\partial x_k}=\sum_{k=1}^n\,\frac{\partial}{\partial x_k}\,.$$
Hence, the given partial differential equation is equivalent to $$\frac{\partial \phi}{\partial y_n}(y_1,y_2,\ldots,y_{n-1},y_n)=0\,,$$
where $$\phi(y_1,y_2,\ldots,y_{n-1},y_n):=f\left(y_1+y_n,y_2+y_n,\ldots,y_{n-1}+y_n,y_n\right)\,.$$  Consequently, $\phi(y_1,y_2,\ldots,y_{n-1},y_n)$ is a function of $y_1,y_2,\ldots,y_{n-1}$.  In other words, there exists a smooth function $\Phi:\mathbb{R}^{n-1}\to \mathbb{R}$ such that
$$f\left(x_1,x_2,\ldots,x_{n-1},x_n\right)=\Phi\left(x_1-x_n,x_2-x_n,\ldots,x_{n-1}-x_n\right)\,.$$
We can check that such $f$ satisfies the condition.  Your nontrivial example for even $n$ arises from taking
$$\Phi(t_1,t_2,\ldots,t_{n-1}):=\exp\left(\sum_{i=1}^{n-1}\,(-1)^{i+1}\,t_i\right)\,.$$
In general, let $\alpha_1,\alpha_2,\ldots,\alpha_n\in\mathbb{R}$ be arbitrary with $\alpha_n\neq 0$.  Then, all differentiable functions $f:\mathbb{R}^n\to\mathbb{R}$ that satisfy the partial differential equation
$$\sum_{i=1}^n\,\alpha_i\,\frac{\partial f}{\partial x_i}(x_1,x_2,\ldots,x_n)=0$$
for every $x_1,x_2,\ldots,x_n\in\mathbb{R}$ take the form
$$f(x_1,x_2,\ldots,x_n)=\Phi\left(\alpha_n\,x_1-\alpha_1\,x_n,\alpha_n\,x_2-\alpha_2\,x_n,\ldots,\alpha_n\,x_{n-1}-\alpha_{n-1}\,x_n\right)\,,$$
where $\Phi:\mathbb{R}^{n-1}\to\mathbb{R}$ is a differentiable function.
A: Note that your equation is equivalent to
$$
(1,1,1,\dots,1)\cdot\nabla f=0
$$
This means that $f$ is constant on lines parallel to $(1,1,1,\dots,1)$; these lines are the characteristics of the partial differential equation.
Thus, we can define $f$ freely on $x_n=0$ and then
$$
f(x_1,x_2,x_3,\dots,x_n)=f(x_1-x_n,x_2-x_n,x_3-x_n,\dots,0)
$$
