Answer
$(4,2)$ and $(0,0)$
Work Step by Step
Step 1. Given the equation of the function $y=\frac{x}{x-2}$, we can find the derivative as $y'=\frac{x-2-x}{(x-2)^2}=\frac{-2}{(x-2)^2}$
Step 2. The tangent line perpendicular to the line $y=2x+3$ will have a slope given by $m=-\frac{1}{2}$
Step 3. Let $y'=m$, we have $\frac{-2}{(x-2)^2}=-\frac{1}{2}$, which gives $(x-2)^2=4$ or $x-2=\pm2$; thus $x=2\pm2$
Step 4. For $x=4$, we have $y=\frac{4}{4-2}=2$ or point $(4,2)$
Step 5. For $x=0$, we have $y=\frac{0}{0-2}=0$ or point $(0,0)$
