Answer: (i) \(\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{1-2t^2}{2t-2t^3}\).

(ii) \(\dfrac{\mathrm{d}^2 y}{\mathrm{d} x^2}=\dfrac{(-1+t^2-2t^4)e^{t^2}}{3}\).

Given \(x=t^2e^{-t^2}\) and \(y=te^{-t^2}\).

(i) Differentiate both with respect to \(t\):

\(\dfrac{\mathrm{d} x}{\mathrm{d} t}=2t e^{-t^2}+t^2(-2t)e^{-t^2}=(2t-2t^3)e^{-t^2}\).

\(\dfrac{\mathrm{d} y}{\mathrm{d} t}=e^{-t^2}+t(-2t)e^{-t^2}=(1-2t^2)e^{-t^2}\).

Then

\(\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{\mathrm{d} y/\mathrm{d} t}{\mathrm{d} x/\mathrm{d} t}=\dfrac{(1-2t^2)e^{-t^2}}{(2t-2t^3)e^{-t^2}}=\dfrac{1-2t^2}{2t-2t^3}.\)

(ii) Differentiate \(\dfrac{\mathrm{d} y}{\mathrm{d} x}\) with respect to \(t\), then divide by \(\dfrac{\mathrm{d} x}{\mathrm{d} t}\):

\(\dfrac{\mathrm{d} y}{\mathrm{d} x}=\dfrac{1-2t^2}{2t-2t^3}=\dfrac{1-2t^2}{2t(1-t^2)}.\)

Using the quotient rule on \(\dfrac{1-2t^2}{2t-2t^3}\),

\(\dfrac{\mathrm{d}}{\mathrm{d} t}\left(\dfrac{\mathrm{d} y}{\mathrm{d} x}\right)=\dfrac{(-4t)(2t-2t^3)-(1-2t^2)(2-6t^2)}{(2t-2t^3)^2}.\)

Simplifying the numerator gives

\((-4t)(2t-2t^3)-(1-2t^2)(2-6t^2)=2+4t^2-4t^4=2(1+2t^2-2t^4).\)

So

\(\dfrac{\mathrm{d}}{\mathrm{d} t}\left(\dfrac{\mathrm{d} y}{\mathrm{d} x}\right)=\dfrac{2(1+2t^2-2t^4)}{(2t-2t^3)^2}.\)

Now

\(\dfrac{\mathrm{d}^2 y}{\mathrm{d} x^2}=\dfrac{\mathrm{d}}{\mathrm{d} t}\left(\dfrac{\mathrm{d} y}{\mathrm{d} x}\right)\cdot \dfrac{\mathrm{d} t}{\mathrm{d} x}.\)

Since \(\dfrac{\mathrm{d} t}{\mathrm{d} x}=\dfrac{1}{\mathrm{d} x/\mathrm{d} t}=\dfrac{1}{(2t-2t^3)e^{-t^2}}\),

\(\dfrac{\mathrm{d}^2 y}{\mathrm{d} x^2}=\dfrac{2(1+2t^2-2t^4)}{(2t-2t^3)^2}\cdot \dfrac{1}{(2t-2t^3)e^{-t^2}}.\)

Simplifying,

\(\dfrac{\mathrm{d}^2 y}{\mathrm{d} x^2}=\dfrac{(-1+t^2-2t^4)e^{t^2}}{3}.\)

So the required second derivative is \(\dfrac{(-1+t^2-2t^4)e^{t^2}}{3}\).