System of differential equations $dx=\frac{dy}{y+z}=\frac{dz}{x+y+z}$ This is the first time I have seen system of differential equations in this form: $$dx=\frac{dy}{y+z}=\frac{dz}{x+y+z}$$
Can you please help me solve it because I don't even know where to start? 

$$\frac{dx}{1}=\frac{dy}{y+z}=\frac{dz}{x+y+z}$$
This system looks like to be involved in solving a PDE with the method of characteristics. The PDE should be :
$$\frac{\partial z(x,y)}{\partial x}+(y+z(x,y))\frac{\partial z(x,y)}{\partial y}=x+y+z(x,y)$$
$\underline{\text{If this supposition is true}}$,
unfortunately the boundary conditions are missing in the wording of the question. 
$$\frac{dy}{y+z}=\frac{dz}{x+y+z}=\frac{dz-dy}{(x+y+z)-(y+z)}=\frac{dz-dy}{x}$$
A first family of characteristics comes from $\quad \frac{dx}{1}=\frac{dz-dy}{x} \quad\to\quad z-y-\frac{x^2}{2}=c_1$
A second family of characteristics comes from $\quad \frac{dx}{1}=\frac{dy}{y+z}=\frac{dy}{y+(y+\frac{x^2}{2}+c_1)}=\frac{dy}{2y+\frac{x^2}{2}+c_1}$
$\frac{dy}{dx}=2y+\frac{x^2}{2}+c_1 \quad\to\quad y=-\frac{c_1}{2}-\frac{x^2}{4}-\frac{x}{4}-\frac{1}{8}+c_2e^{2x}$
$y+\frac{z-y-\frac{x^2}{2}}{2}+\frac{x^2}{4}+\frac{x}{4}+\frac{1}{8}=c_2e^{2x} \quad\to\quad (4z+4y+2x+1)e^{-2x}=8c_2$
The general solution of the PDE is expressed on the form of an implicit equation :
$$\Phi\left((2z-2y-x^2)\:,\:(4z+4y+2x+1)e^{-2x} \right)=0$$
where $\Phi$ is any function of two variables (to be determined according to some boundary conditions).
$\underline{\text{If the above supposition is false}}$ :
Then, $z$ is function of $x$ only, that is $z(x)$ instead of $z(x,y)$.
The system becomes : 
$$1=\frac{y'}{y+z}=\frac{z'}{x+y+z}$$
Following the same calculus, the result is :
$z(x)=y+\frac{x^2}{2}+c_1$
$ y(x)=-\frac{c_1}{2}-\frac{x^2}{4}-\frac{x}{4}-\frac{1}{8}+c_2e^{2x}$
$$\begin{cases}
y(x)=-\frac{c_1}{2}-\frac{x^2}{4}-\frac{x}{4}-\frac{1}{8}+c_2e^{2x} \\
z(x)=\frac{c_1}{2}+\frac{x^2}{4}-\frac{x}{4}-\frac{1}{8}+c_2e^{2x}
\end{cases}$$
