Consider $\dot{x}=f(x)$ where differentiation is with respect to $t$. Suppose there exist $x_0>0$, such that $f\left(x_0\right)=0$, and $\epsilon>0$ such that $$ \left|\int_{x_0}^{x_0+\epsilon} \frac{d \xi}{f(\xi)}\right|<\infty . $$ Let $y=1 / x$ and $s=-t$. Find the ODE for $y$ with respect to $s$ and prove that there exists $\delta \in(0, \infty)$ such that when $y(0)=\delta$, the solution to the ODE for $y$ loses uniqueness in finite positive time.
My try: Now once I use the substitution we get $\frac{\dot{y}}{y^2}=f\left(\frac 1y\right)$ and $$ \left|\int_{x_0}^{x_0+\epsilon} \frac{d \xi}{f(\xi)}\right|<\infty . $$ transforms $$ \left|\int_{\frac 1{x_0}}^{\frac 1{x_0+\epsilon}} \frac{d \eta}{\eta^2f(\frac 1{\eta})}\right|<\infty . $$ if we put $\xi=\frac 1{\eta}$. Now we name, $\frac 1{x_0}=y_0$ and $\frac 1{x_0+\epsilon}=y_0+\theta$ then
$$ \left|\int_{\frac 1{x_0}}^{\frac 1{x_0+\epsilon}} \frac{d \eta}{\eta^2f(\frac 1{\eta})}\right|=\left|\int_{y_0}^{y_0+\theta} \frac{d \eta}{\eta^2f(\frac 1{\eta})}\right|<\infty . $$ so there exists $y_1 \in (y_0,y+0+\theta)$ s.t $y_1f(\frac1{y_1})\neq 0$ next we can define an inverse of $F$ in a neighbourhood of $y_1$ where $F(y(s))= \int_{y_0}^{y}\frac{d \eta}{\eta^2f(\frac 1{\eta})}$. Am I correct? Now I am not getting any way out. Please help.