The system is:
\begin{equation*}
x''=\frac{-\mu x}{(\mu x^2 + y^2)^{3/2}}
\end{equation*}
\begin{equation*}
y''=\frac{-y}{(\mu x^2 + y^2)^{3/2}}
\end{equation*}
With $\mu>1$ a constant adding anisotropy to the otherwise newtonian system.
With energy function:
\begin{equation*}
E: \mathbb{R}^4:\rightarrow \mathbb{R}
\end{equation*}
\begin{equation*}
(v_{x},v_{y},x,y)\mapsto \frac{1}{2}\sqrt{v_{x}^2+v_{y}^2}-
\frac{1}{\sqrt{\mu x^{2}+y^{2}}}
\end{equation*}
Which I have reparametrized, reducing the velocity dimension to 1 in order to plot level curves in configuration space (for only negative values of energy) as:
\begin{equation*}
E(V,x,y)=\frac{|V|^2}{2}-
\frac{1}{\sqrt{\mu x^{2}+y^{2}}}
\end{equation*}
Squaring the norm of velocity for smoothness (I believe).
The constraints on the solution that i must prove lies on the $x$ or $y$ axes of configuration space are twofold: 1. Solution must meet the $V=0$ curve and; 2. The solution must be a straight-line collision ejection orbit.
I believe the first condition implies that, after solving the $E$ funciton for $\frac{|V|^2}{2}$ and setting to zero implies: \begin{equation*} \sqrt{\mu x^2+y^2}=\frac{-1}{h} \end{equation*}
My trouble is with the second condition as I am not quite sure what it means (particularly the "straight-line" portion, which was not defined) but from what i have gathered from Hirsch-Smale and Devaney chapter 13.4, this means that evaluating the limit as $x\rightarrow 0$ means $y\rightarrow \infty$ and the same for $y\rightarrow 0$ and $x\rightarrow \infty$ or solutions go to infinity in finite time in both directions. I think this then should imply that solutions "lie on" the x and y axes. Velocity or the z direction has been set to zero and we are sending either coordinate to 0, yielding a one dimensional explosion along one of the two position coordinates in configuration space. Does this make sense? Any clarification or resources on the meaning of condition 2 would be appreciated.
Firstly, the level sets of the function $E_1(V,x,y) = \frac{|V|}{2} - \frac{1}{\sqrt{\mu x^2+y^2}}$ do not coincide with the level sets of the function $E_2(V,x,y) = \frac{|V|^2}{2} - \frac{1}{\sqrt{\mu x^2+y^2}}$, because $E_1$ and $E_2$ are not the same function.
Secondly, the first energy function you defined is not conserved by the system, as \begin{equation} \frac{\text{d}E_1}{\text{d} t} = \frac{\mu x x' + y y'}{(\mu x^2 + y^2)^{\frac{3}{2}}}\left(1-\frac{1}{2}\frac{1}{\sqrt{x'^2+y'^2}}\right) \neq 0. \end{equation} However, $E_2$ is conserved, yielding a proper energy function for the system.
Thirdly, the condition that the solution you seek meets the $V=0$-curve is not equivalent with the condition that it entirely coincides with the $V=0$-curve, which is what you assumed.
Forthly, the solution being a 'straight line collision orbit' is highly coordinate dependent. I assume (and rightly so, I believe) that your $x$ and $y$ axes are the Cartesian coordinate axes for the physical system. In those coordinates, a straight line obeys the relation $y = a x + b$. Assuming that the collision should be with the mass in the centre of the coordinate system, this means that this straight line should go through the origin, i.e. that $b=0$. Note that the only straight line not covered by the relation $y=a x$ is the line $x=0$, which coincides with the $y$-axis.