I was presented with the equation:
$\begin{equation} \begin{split} \frac{dx}{x(x+y)} = \frac{dy}{y(x+y)} = \frac{dz}{(x-y)(2x+2y+z)} \end{split} \end{equation}$
So, we immediately know that the solution must be in this form:
$F(c_1,c_2) =0$.
Taking the the first two terms:
$\begin{equation} \begin{split} \frac{dy}{dx} = \frac{y(x+y)}{x(x+y)} = \frac{y}{x} \end{split} \end{equation}$
$\begin{equation} \begin{split} \int_{}{\frac{1}{x}} \ dx =\int_{}{\frac{1}{y}} \ dy \end{split} \end{equation}$
Then,
$\begin{equation} \begin{split} \ln {\frac{x}{y}} => \frac{x}{y} = c_1 \end{split} \end{equation}$
The solution: $\frac{x}{y} = c_1$
Next,
$\begin{equation} \begin{split} \frac{dx - dy}{(x-y)(x+y)}= \frac{dz}{(x-y)(2x+2y+z)} \end{split} \end{equation}$
Finally subtracting and multiplying by a factor of $(x-y)$ (on both sides),
$\begin{equation} \begin{split} \frac{dx-dy-dz}{(-x-y+z)} = 0 \end{split} \end{equation}$
The solution is: $\frac{x}{yz^2} =c_2$ So, The solution is $F(\frac{x}{y},\frac{x}{yz^2}) =0$
Am I correct?