$x = \frac{{1 + \log t}}{{{t^2}}},y = \frac{{3 + 2\log t}}{t}$

and $y = \frac{{3 + 2\log t}}{t}$

therefore,$\frac{{dx}}{{dt}} = \frac{{{t^2} \cdot \frac{d}{{dt}}(1 + \log t) - (1 + \log t) \cdot \frac{d}{{dt}}{t^2}}}{{{{\left( {{t^2}} \right)}^2}}}$

$= \frac{{{t^2} \cdot \frac{1}{t} - (1 + \log t) \cdot 2t}}{{{t^4}}} = \frac{{t - (1 + \log t) \cdot 2t}}{{{t^4}}}$
$= \frac{t}{{{t^4}}}[1 - 2(1 + \log t)] = \frac{{ - 1 - 2\log t}}{{{t^3}}}$ ……(i)

and $\frac{{dy}}{{dt}} = \frac{{t \cdot \frac{a}{{dt}}(3 + 2\log t) - (3 + 2\log t) \cdot \frac{a}{{dt}}t}}{{{t^2}}}$
$= \frac{{t \cdot 2 \cdot \frac{1}{t} - (3 + 2\log t) \cdot 1}}{{{t^2}}}$

$= \frac{{2 - 3 - 2\log t}}{{{t^2}}} = \frac{{ - 1 - 2\log t}}{{{t^2}}}$ …….(ii)

therefore,$\frac{{dy}}{{dx}} = \frac{{dy/dt}}{{dx/dt}} = \frac{{ - 1 - 2\log t/{t^2}}}{{ - 1 - 2\log t/{t^3}}} = t$

After simplification we get : $\frac{dy}{dx} = t$