If a function is represented parametrically by the equations $x=\frac{1+{\log }_{e}t}{t^2};y=\frac{3+2{\log }_{e}t}{t},$ then which of the following statement are true

$x=\frac{1+{\log }_{e}t}{t^2};y=\frac{3+2{\log }_{e}t}{t}\Rightarrow \frac{dy}{dx}=\frac{dy/dt}{dx/dt}$

$=\frac{\frac{t\left(\frac{2}{t}\right)-\left(3+2{\log }_{e}t\right)}{t^2}}{\frac{t^2\left(\frac{1}{t}\right)-\left(1+{\log }_{e}t\right)2t}{t^4}}=\left(\frac{-1-2{\log }_{e}t}{-1-2{\log }_{e}t}\right)t=t$

Eliminating ${\log }_e$ t term from y , we get

$y=\frac{1+2t^{2}x}{t}=\frac{1+2{\left(y^{\mathrm{\prime }}\right)}^{2}x}{y^{\mathrm{\prime }}}\text{ or }y{y}^{\mathrm{\prime }}=1+2x{\left(y^{\mathrm{\prime }}\right)}^2$

$\text{ or }y{y}^{\prime \prime }+{\left(y^{\mathrm{\prime }}\right)}^2=4x{y}^{\mathrm{\prime }}{y}^{\prime \prime }+2{\left(y^{\mathrm{\prime }}\right)}^2\text{ or }yy"=4xy\cdot y^{\prime \prime }+{\left(y^{\mathrm{\prime }}\right)}^2$