Stability Analysis Near Fixed Point with Suspected Centers $$
\begin{cases}
 \dot{x} = -y\\
 \dot{y} = x - \frac{1}{2}x^2 + \frac{1}{2} y^2 \\
\end{cases}
$$
I'm working on this system right now, and am unsure how to analyze it near the fixed point $(0,0)$. 
The eigenvalues of the linearized system have non-zero real part, so linearization is inconclusive. I then plotted it in Python and some online tools to see what I should expect, and consistently saw a continuum of centers near the origin. I'm used to this signalling a Hamiltonian system, but it isn't here. Moreover, since the periodic orbits are not limit cycles, Poincaré-Bendixson theorem isn't useful to my knowledge. 
I'm thinking that there should be an invariant of the system that describes these orbits. That is, some $G(x, y)$ such that $\dot{G} = 0$, or at least a (not strict) Lyapunov Function. But guessing the form is proving difficult. The orbits seem to collapse into circles near the origin, so I suspected that it would be something like:
$$G(x, y) = x^2 + y^2 + f(x, y)$$
Where $f(x, y) = o(x^2 + y^2)$. This will be positive in a neighborhood of the origin (excluding the origin). With $f(x, y) = -x^3$, we get:
$$\dot{G} = (2x - x^2)(-y) + 2y(x - \frac{1}{2}x^2 + \frac{1}{2}y^2 ) = y^3 $$
Which doesn't seem like a bad start, but isn't there yet. I believe that $f(x, y)$ will require both $x$ and $y$ dependence, but again I'm not sure how yet.
 A: Solution
Here's the exact solution of the ODE system 
$$x'(t) = -y(t)$$
$$y'(t) = x(t) -\frac{1}{2}(x(t)^2-y(t)^2)$$
in terms of trajectories $y=y(x)$, i.e. we provide the exact form of the invariant $G(x,y)$ discused in the OP.
The transformation $x = u+v$, $y=u-v$, (backwards $u=\frac{1}{2}(x+y)$, $v=\frac{1}{2}(x-y)$)
gives the ODEs
$$u'+v'=-u+v$$
$$u'-v'=u+v-2 u v$$
or, adding and subtracting,
$$2u' = 2 v - 2 u v= 2 v(1-u)$$
$$2v' = -2 u + 2 u v = -2 u(1-v)$$
Dividing these gives
$$du \frac{u}{1-u} + dv \frac{v}{1-v}=0$$
which can be integrated with the result
$$-u - \log(1-u) - v - \log(1-v) = const$$
Transforming back gives
$$\log (2-x-y)+\log (2-x+y)+x=c\tag{1}$$
This is the invariant looked for in the OP.
Solving for $y$ shows that the trajectories are given by the simple formula
$$y =\pm\sqrt{(x-2)^2-a e^{-x}}\tag{2}$$
where $a$ is a constant.
Here's a plot of the trajectories for $a=2$.

The complete stream plot of the ODE in the given region about the critical points $p_1=(0,0)$ an $p_2=(2,0)$ is:

Discussion
The closed part of the trajectory is an interesting curve which looks like a rounded triangle. I don't know if it has a specific name.
For $x\to 0$ and $a\lt 4$ formula $(2)$ becomes
$$y\simeq \pm(\frac{\sqrt{4-a} (a+4)}{2 a}-\frac{a \left(\frac{8}{a}+x-2\right)^2}{8 \sqrt{4-a}})\tag{3}$$
These are two parabolas which embrace the exact curve in a certain range but deviate strongly in others as is shown in the next graph

For $|x|<<1$ we have from $(2)$ by expanding $e^{-x}$
$$y^2  \simeq (x^2-4x+4)-a(1-x+\frac{1}{2}x^2)+O(x^3)\\\
=-\frac{1}{2}(a-2)x^2 - x(4-a) + (4-a)$$
and for $a\simeq 4$
$$x^2+y^2  \simeq  (4-a)\tag{4}$$
which is the equation of a circle of radius $\sqrt{4-a}$. This result meets the original expectations. (Thanks to infinitylord for pointing this out.)
The meaning of the parameter $a$ is found from $(2)$
$$a = e^{x}\left((x-2)^2-y^2\right)\tag{5a}$$
by applying the initital conditions.
Letting $x(t=0)=0$ and $y(0)=y_{0}$ we get
$$a = 4-y_0^2\tag{5b}$$
Hence the radius in $(4)$ is equal to $y_0$.
On the other hand, for $y=0$ we have from $(2)$
$$(x-2)^2=a e^{-x}\tag{5c}$$
The next plot shows the solution of $(5c)$ in inverse form $a(x)$ 

Finally, for $x=2$ we have $-y^2 = a e^{-2}$ or
$$a = - (e y_0)^2\tag{5d}$$
In polar coordinates $(2)$ bcomes
$$r^2 \sin(\phi)^2= (r \cos(\phi) -2)^2 + a e^{- r \cos(\phi)}\tag{6}$$
and here are polar plots of this relation for some values of $a$

The time behaviour of the system can be found by integrating
$$x'(t) = y(x)$$
Specifically, the period for the closed trajectory is given by
$$T(a) = 2 \int_{x_{min}}^{x_{max}}\frac{1}{\sqrt{(x-2)^2- a e^{-x}}}\,dx\tag{7}$$
Here $x_{min}(a)$ and $x_{max}(a)$ are the zeroes of the radicand, c.f. $(5c)$.
