$\begin{align}  & x=\frac{2+t}{1+2t} \\ & \frac{dx}{dt}=\frac{(1+2t)\tfrac{d}{dt}(2+t)-(2+t)\tfrac{d}{dt}(1+2t)}{{{(1+2t)}^{2}}} \\ & \frac{dx}{dt}=\frac{(1+2t)(1)-(2+t)(2)}{{{(1+2t)}^{2}}} \\ & \frac{dx}{dt}=\frac{1+2t-4-2t}{{{(1+2t)}^{2}}}=\frac{-3}{{{(1+2t)}^{2}}} \\ & \frac{dt}{dx}=-\frac{{{(1+2t)}^{2}}}{3} \\ & y=\frac{3+2t}{t} \\ & \frac{dy}{dt}=\frac{t\tfrac{d}{dt}(3+2t)-(3+2t)\tfrac{d}{dt}(t)}{{{t}^{2}}} \\ & \frac{dy}{dt}=\frac{t(2)-(3+2t)(1)}{{{t}^{2}}}=\frac{2t-3-2t}{{{t}^{2}}} \\ & \frac{dy}{dt}=-\frac{3}{{{t}^{2}}} \\ & \text{when }x=0 \\ & \frac{2+t}{1+2t}=0 \\ & t=-2 \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dx}\left( \frac{dy}{dx} \right)=\frac{d}{dx}\left( \frac{{{(1+2t)}^{2}}}{{{t}^{2}}} \right) \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{d}{dt}\left[ \frac{{{(1+2t)}^{2}}}{{{t}^{2}}} \right]\frac{dt}{dx} \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{{{t}^{2}}\tfrac{d}{dt}{{(1+2t)}^{2}}-{{(1+2t)}^{2}}\tfrac{d}{dt}({{t}^{2}})}{{{({{t}^{2}})}^{2}}}\cdot \frac{{{(1+2t)}^{2}}}{-3} \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{{{t}^{2}}[2(1+2t)\cdot 2]-{{(1+2t)}^{2}}(2t)}{-3{{t}^{4}}}\cdot {{(1+2t)}^{2}} \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}={{(1+2t)}^{2}}\left[ \frac{4{{t}^{2}}(1+2t)-2t{{(1+2t)}^{2}}}{-3{{t}^{4}}} \right] \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=2t{{(1+2t)}^{3}}\left[ \frac{2-(1+2t)}{-3{{t}^{4}}} \right] \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{2{{(1+2t)}^{3}}(2t-1)}{3{{t}^{3}}} \\ & \text{When }x=0,t=-2 \\ & \frac{{{d}^{2}}y}{d{{x}^{2}}}=\frac{2{{(1-4)}^{3}}(-4-1)}{-24}=\frac{270}{24}=\frac{45}{4} \\\end{align}$