$x=\frac{1+\log t}{t^{2}}, \quad y=\frac{3+2 \log t}{t}$
On differentiating both the given parametric functions w.r.t. t, we have
$\frac{d x}{d t}=\frac{t^{2} \cdot \frac{d}{d t}(1+\log t)-(1+\log t) \cdot \frac{d}{d t}\left(t^{2}\right)}{t^{4}}$
$=\frac{t^{2} \cdot\left(\frac{1}{t}\right)-(1+\log t) \cdot 2 t}{t^{4}}=\frac{t-(1+\log t) \cdot 2 t}{t^{4}}$
$=\frac{t[1-2-2 \log t]}{t^{4}}=\frac{-(1+2 \log t)}{t^{3}}$
$y=\frac{3+2 \log t}{t}$
Next,
$\frac{d y}{d t}=$ $\frac{t \cdot \frac{d}{d t}(3+2 \log t)-(3+2 \log t) \cdot \frac{d}{d t}(t)}{t^{2}}$
$=\frac{t(2 / t)-(3+2 \log t) \cdot 1}{t^{2}}$
$=\frac{2-3-2 \log t}{t^{2}}=\frac{-(1+2 \log t)}{t^{2}}$
$\therefore \frac{d y}{d x}=\frac{d y / d t}{d x / d t}=\frac{\frac{-(1+2 \log t)}{t^{2}}}{\frac{-(1+2 \log t)}{t^{3}}}=\frac{t^{3}}{t^{2}}=t$
Thus, $\frac{d y}{d x}=t$.
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