Limit cycle within $\frac{1}{4}Show that the system
\begin{align}
    x'&=-y+x(1-2x^2-3y^2)\nonumber\\
    y'&=x+y(1-2x^2-3y^2)\nonumber
\end{align}
has a limit cycle in $\frac{1}{4}<r<1$.
Here's what I've done so far:
\begin{align}
    xx'&=-xy+x^2(1-2x^2-3y^2)\nonumber\\
    yy'&=xy+y^2(1-2x^2-3y^2)\nonumber
\end{align}
\begin{align}
    xx'+yy'&=-xy+x^2(1-2x^2-3y^2)+xy+y^2(1-2x^2-3y^2)\nonumber\\
   \frac{1}{2}(x^2+y^2)'&=x^2(1-2x^2-3y^2)+y^2(1-2x^2-3y^2) \nonumber\\
   \frac{1}{2}(x^2+y^2)'&=(x^2+y^2)(1-2x^2-3y^2) \nonumber
\end{align}
Knowing that $r^2=x^2+y^2$:
\begin{align}
   \frac{1}{2}(x^2+y^2)'&=(x^2+y^2)(1-2x^2-3y^2) \nonumber\\
   \frac{1}{2}(r^2)'&=(r^2)(1-2x^2-3y^2) \nonumber
\end{align}
I'm now stuck here and I'm not really sure what to do next because of this part $(1-2x^2-3y^2)$.

Noting that on $r=1$,
$$ \frac{1}{2}(r^2)'=(r^2)(1-2x^2-3y^2)=r^2(1-2r^2-y^2)<0 $$
and on $r=\frac14$,
$$ \frac{1}{2}(r^2)'=(r^2)(1-2x^2-3y^2)=r^2(1-3r^2+y^2)>0 $$
by the Poincare-Bendixson Theorem, there is a closed trajectory lying in $\frac14<r<1$.
