System of first-order ordinary differential equations solve for $x,y,z$:
$$\frac{dx}{x^{2}+a^{2}}=\frac{dy}{xy-az}=\frac{dz}{xz+ay}$$
please give a hint. I am not able to formulate the steps required to proceed solving this one.

We have 
$$
\begin{align}
\frac{dy}{xy-az} & = \frac{dz}{xz+ay}\\
\frac{dy/y}{x-a (z/y)} & = \frac{dz/z}{x+a (y/z)}
\end{align}
$$
This gives a motivation to let $z = ky$ where $k$ is a constant.
$$
\begin{align}
\frac{dy/y}{x-a k} & = \frac{dy/y}{x+a/k}
\end{align}
$$
This gives us that $k = \pm i$. Let $k=i$.
This gives us that 
$$\begin{align}
\frac{dx}{x^2 + a^2} & = \frac{dy/y}{x - ia}\\
\frac{dx}{x + ia} & = \frac{dy}{y}\\
y & = c(x + ia)
\end{align}
$$
Hence, we get
$$
\begin{align}
z & = ic(x+ia)\\
y & = c(x+ia)
\end{align}
$$
and
$$
\begin{align}
z & = -ic(x-ia)\\
y & = c(x-ia)
\end{align}
$$
I don't know to justify my motivation why I chose $z = ky$ instead of $z=k(y)y$.
