I want to show that for the system $\dot{x}=x^2, \dot{y}=y^2$,any solutions starting in the 3rd quadrant not including 0, converge uniformly to the origin. For an initial point $(x_0,y_0)$, (note both $x_0$ and $y_0$ are negative as we are considering in the 3rd quadrant), we have solutions to the system as $$x(t)=\frac{1}{\frac{1}{x_0}-t}, y(t)=\frac{1}{\frac{1}{y_0}-t}$$ Let's just consider the x(t) (as y(t) will be similar). As $x_0$ is negative, let us say that $x_0<-\epsilon$, for some $\epsilon>0$.
To show the uniform convergence, I want to show that
$$\sup_{x_0 \leq -\epsilon} |x(t)| \rightarrow 0 \text{ as } t\rightarrow \infty$$ $$=\sup_{x_0 \leq -\epsilon} \frac{1}{t-\frac{1}{x_0}}$$
But I am not sure where to go from here. Could I write $$=\sup_{x_0 \leq \epsilon} \frac{x_0}{x_0t-1} \leq \frac{-\epsilon}{-\epsilon t -1} \leq \frac{\epsilon}{\epsilon t+1} \rightarrow 0 \text{ as } t \rightarrow \infty$$? Any help is appreciated.
If $x_0 \leqslant -\epsilon < 0$ then
$$\left| \frac{1}{ 1/x_0-t}\right| = \left| \frac{-1}{ t - 1/x_0}\right| =\frac{1}{t + 1/|x_0|} \leqslant \frac{1}{t}$$
Given $\epsilon_0 > 0$, if $t > 1/\epsilon_0$ then for all $x_0 \leqslant -\epsilon$ we have
$$\left| \frac{1}{1/x_0 - t}\right| < \epsilon_0.$$
Therefore, convergence to $0$ is uniform.