How to interpret the differential equation (z+x)dx + (x+z)dy + (x+y)dz = 0? Since $\frac{dy}{dx}$ is not a ratio but a limit, the differential equation 
$\frac{d^2y}{dx^2} -3 \frac{dy}{dx} + 2y = 0$ can be interpreted as a function $y$ whose derivatives first order and second order satisfy the equation and not like differentials $dx, dy$ satisfy the equation.
But how to interpret a differential equation of the form $(z+x)dx + (x+z)dy + (x+y)dz = 0$ and also is there any rigorous way of expressing the equation above. Since a derivative must be expressed as $\frac{dy}{dx} = \lim_{h\to 0} \frac{f(x+h)- f(x)}{h}$?
Thanks in advance.
 A: Think of it in finite terms. Swap the differential $d$ for a finite diference $\Delta$:
$$
(z+x) \Delta x + (x+z) \Delta y + (x+y)\Delta z = 0
$$
This can be written in vectorial form as
$$
\begin{pmatrix} z+x \\ x+z \\ x+y \end{pmatrix} \cdot \begin{pmatrix} \Delta x \\ \Delta y \\ \Delta z\end{pmatrix} = 0
$$
What is it saying then?
Suppose $x = x(t)$, $y = y(t)$ and $z = z(t)$, i.e., the triplet $(x,y,x)$ defines a curve in space; then,
\begin{align}
\Delta x &= x(t + \Delta t) - x(t),\\
\Delta y &= y(t + \Delta t) - y(t),\\
\Delta z &= z(t + \Delta t) - z(t),\\
\end{align}
So, the vectorial equation can be written as
$$
\begin{pmatrix} z(t)+x(t) \\ x(t)+z(t) \\ x(t)+y(t) \end{pmatrix} \cdot \begin{pmatrix} \frac{x(t + \Delta t) - x(t)}{\Delta t} \\ \frac{y(t + \Delta t) - y(t)}{\Delta t} \\ \frac{z(t + \Delta t) - z(t)}{\Delta t}\end{pmatrix} \Delta t = 0
$$
If the change in $t$ is infinitesimal, then
$$
\begin{pmatrix} z(t)+x(t) \\ x(t)+z(t) \\ x(t)+y(t) \end{pmatrix} \cdot \begin{pmatrix} x'(t) \\ y'(t) \\ z'(t) \end{pmatrix} d t = 0
$$
And, of course, $dt$ is as small as you like, but not zero (remember your epsilons and deltas), so
$$
\begin{pmatrix} z(t)+x(t) \\ x(t)+z(t) \\ x(t)+y(t) \end{pmatrix} \cdot \begin{pmatrix} x'(t) \\ y'(t) \\ z'(t) \end{pmatrix} = 0
$$
Now, what on Earth is the vector on the right?
$\color{blue}{\text{The tangent vector to the curve!}}$
So, your differential equation is telling:

You need to find a curve in space such that its tangent vector is orthogonal to the vector defined by
  $$\mathbf{v}(t) = [z(t) + x(t)]\mathbf{i} + [x(t) + z(t)]\mathbf{j} + [x(t) + y(t)]\mathbf{k}$$
  for all values of the parameter $t$.

If $\mathbf{v}(t)$ is orthogonal to $\mathbf{T}$, then it inhabits in the plane defined by $\mathcal{P} := \{\mathbf{v} \in \mathbb{R}^3\,|\, \mathbf{v}\cdot \mathbf{T} = 0\}$. In other words,
$$
\mathbf{N}(t) = \mathbf{T}'(t) = \lambda \mathbf{v}(t)
$$
where $\mathbf{N} \in \mathcal{P}$ and $\lambda$ is a constant. 
Of course, as stated, the solution to the problem isn't unique.
But, why?
Well, suppose you have a function $u = u(x,y,z)$ and, as stated, $x = x(t)$, $y = y(t)$ and $z = z(t)$. Then,
$$
\frac{d u}{d t} = u_x x'(t) + u_y y'(t) + u_z z'(t) = 0.
$$
Another reading to the DE is:

You are looking for a function $u(x,y,z)$ such that its gradient is collinear to $\mathbf{v}$:
  $$\mu\nabla u = [z + x]\mathbf{i} + [x + z]\mathbf{j} + [x + y]\mathbf{k}$$
  where $\mu$ is a constant.

So, there is an infinte number of functions that satisfy this property. Even if you focus only on the ones that pass trough a curve such that there is uniqueness for $u$, an infinite number of this curves can be constructed, ensuring lack of uniqueness for the orignial DE (even with initial conditions for $x$, $y$ and $z$ at some $t_0$).
