If $x,y:\mathbb{R}_+\rightarrow\mathbb{R}$ are real analytic functions satisfying $$2\dot{x}x-2\dot{y}y=y\dot{x}-x\dot{y}$$
over $\mathbb{R}_+$, where $\dot{x}$ means $\frac{d}{dt}x(t)$ and similar for $\dot{y}$. In addition, $x(t)y(t)\to 1$ as $t\to\infty$.
Is it possible that $x(t)$ or $y(t)$ is unbounded, i.e., there exists a sequence $(t_k)_{k\in\mathbb{N}}$ such that $x(t_k)\to\infty$ or $y(t_k)\to\infty$ where $t_k\to\infty$ as $k\to\infty$?
I know how to deal with it if the RHS of the displayed equation is $0$, i.e., $$2\dot{x}x-2\dot{y}y=0$$ In this case, I can integrate on both sides and obtain $x^2-y^2=c$ over $\mathbb{R}_+$ for some constant $c$. This together with $x(t)y(t)\to 1$ yields the fact that both $x(t)$ and $y(t)$ are bounded, i.e., there is no such sequence $(t_k)_{k\in\mathbb{N}}$.
Additional information:
For some reason, I also know $\dot{x}$ and $\dot{y}$ are in $L^2(\mathbb{R}_+,\mathbb{R})$, i.e., $$\int_0^\infty \dot{x}(t)^2+\dot{y}(t)^2\;dt<\infty.$$
The differential equation is $$ 2x\dot{x}-2y\dot{y}=y\dot{x}-x\dot{y}\tag{1} , $$ where $xy\to 1$ as $t\to\infty.$ Let $1/y=x+\delta,$ where $\delta\to 0$ as $t\to\infty,$ so that on differentiating $y$ with respect to $t,$ $$\dot{y}=-y^2(\dot{x}+\dot{\delta})=-\dfrac{\dot{x}+\dot{\delta}}{\left(x+\delta\right)^2}\, .\tag{2}$$ Now assume that $x$ is unbounded and $x\to \infty$ as $t\to\infty,$ Then, on substituting for $1/y$ into the differential equation $(1),$ we get $$ 2x\dot{x}+\dfrac{2(\dot{x}+\dot{\delta})}{\left(x+\delta\right)^2}\dfrac{1}{x+\delta}=\dfrac{\dot{x}}{x+\delta}+\dfrac{x(\dot{x}+\dot{\delta})}{\left(x+\delta\right)^2}\,. $$ Multiplying through by $\left(x+\delta\right)^3$ gives $$ 2\dot{x}x\left(x+\delta\right)^3+2(\dot{x}+\dot{\delta})=\dot{x}\left(x+\delta\right)^2 +x(\dot{x}+\dot{\delta})\left(x+\delta\right)\,.$$ We have assumed that $x$ is unbounded, so that $x\to\infty$ as $t\to \infty.$ Then as $x\to\infty,\ y\to 0$ and $\delta\to 0$ as $t\to \infty.$ Neglecting $\delta$ compared with $x$ in the preceding equation gives $$ 2\dot{x}x^4+2(\dot{x}+\dot{\delta})=\dot{x}x^2+x^2(\dot{x}+\dot{\delta})\,, $$ or, on collecting terms, $$ 2x^2\dot{x}(x^2-\tfrac{1}{2})=(\dot{x}+\dot{\delta})(x^2-2)\,. $$ Since, by assumption $x^2\gg2>1/2,$ we get $$ 2x^4\dot{x}\approx x^2\dot{x}+\dot{\delta}x^2\quad\text{or}\quad 2x^2\dot{x}(x^2-1/2)\approx \dot{\delta}x^2\,, $$ or finally, by assuming that $x$ is unbounded, we arrive at the relation $$ 2\dot{x}x^2\approx \dot{\delta}\,.\tag{3} $$ This gives, on integration, $$2x^3\approx 3\delta+\text{const}\, ,$$ which is a contradiction, because $x\to\infty\implies \delta\to\infty$ and we require that $\delta\to 0$ as $t\to\infty.$ Therefore there is no such sequence, $(t_k)_{k\in\mathbb{N}}.$
Edit
The differential equation may be written as $$ \frac{\rm d}{{\rm d}t}\left(y^2\left(\frac{x^2}{y^2}-1\right)\right)=y^2\frac{\rm d}{{\rm d}t}\left(\frac{x}{y}\right)\,. $$ Putting $z=x/y$ then gives $$ \frac{\rm d}{{\rm d}t}\left[y^2(z^2-1)\right]=y^2\frac{{\rm d}z}{{\rm d}t}\, , $$ or $$ (z^2-1)\frac{{\rm d}y^2}{{\rm d}t}+y^2\frac{{\rm d}(z^2-1)}{{\rm d}t}=y^2\frac{{\rm d}z}{{\rm d}t}\, . $$ On dividing by $y^2(z^2-1)$ we get $$ \frac{{\rm d}y^2}{y^2}+\frac{{\rm d}(z^2-1)}{z^2-1}=\frac{{\rm d}z}{z^2-1}\, , $$ and on integration $$ \log(y^2)+\log(z^2-1)=\log\sqrt{\frac{z-1}{z+1}}+\text{const}\, , $$ or $$ \log\left((x^2-y^2)\sqrt{\frac{x+y\vphantom{Y^{p}}}{x-y}}\,\,\right)=\text{const}\, . $$ Since the initial conditions are $x=x_{0}$ and $y=y_{0}$ at $t=0,$ we have $$ \log\left((x^2-y^2)\sqrt{\frac{x+y\vphantom{Y^{p}}}{x-y}}\,\,\right)=\log\left((x_{0}^2-y_{0}^2)\sqrt{\frac{x_{0}+y_{0}\vphantom{Y^{p}}}{x_{0}-y_{0}}}\,\,\right)\, .\tag{1} $$
If $x$ is unbounded we require $y\to 0$ with $xy \to 1$ as $t\to\infty.$ Then from $(1),$ as $x$ becomes large and $y$ approaches zero, we obtain $$ \log(x^2)=\text{const}\, , $$ which is a contradiction, so $x$ and $y$ cannot become unbounded if the asymptotic constraint $xy\to 1$ as $t\to\infty$ is to be satisfied.
@William, the OP, has noted that if $z(t)=1$ at some time $t,$ then $\dot{z}=0$ and $z(t)$ will be always $1.$