Answer: (a) \(1-\operatorname{sech}^2 t=\tanh^2 t\).

(b) \(\frac{\mathrm{d}y}{\mathrm{d}x}=-4\operatorname{sech}^2 t\).

(c) The point is \(\left(\frac{8}{25}+\ln\frac{3}{5},\;\frac{881}{625}\right)\).

(a) Using the exponential definitions,

\(\operatorname{sech} t=\frac{2}{e^t+e^{-t}}\), \(\tanh t=\frac{e^t-e^{-t}}{e^t+e^{-t}}\).

Then

\(1-\operatorname{sech}^2 t=1-\left(\frac{2}{e^t+e^{-t}}\right)^2\)

\(=\frac{(e^t+e^{-t})^2-4}{(e^t+e^{-t})^2}\)

\(=\frac{e^{2t}+2+e^{-2t}-4}{(e^t+e^{-t})^2}\)

\(=\frac{e^{2t}-2+e^{-2t}}{(e^t+e^{-t})^2}\)

\(=\frac{(e^t-e^{-t})^2}{(e^t+e^{-t})^2}\)

\(=\left(\frac{e^t-e^{-t}}{e^t+e^{-t}}\right)^2=\tanh^2 t\).

So \(1-\operatorname{sech}^2 t=\tanh^2 t\).

(b) Given

\(x=\frac12\tanh^2 t+\ln(\operatorname{sech} t), \qquad y=1+\tanh^4 t\).

Differentiate with respect to \(t\):

\(\frac{\mathrm{d}y}{\mathrm{d}t}=4\tanh^3 t\,\operatorname{sech}^2 t\).

Also,

\(\frac{\mathrm{d}x}{\mathrm{d}t}=\frac12\cdot 2\tanh t\,\operatorname{sech}^2 t+\frac{1}{\operatorname{sech} t}\cdot \frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech} t)\).

Since \(\frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech} t)=-\operatorname{sech} t\tanh t\), this gives

\(\frac{\mathrm{d}x}{\mathrm{d}t}=\tanh t\,\operatorname{sech}^2 t-\tanh t\)

\(=\tanh t(\operatorname{sech}^2 t-1)\).

Using part (a), \(\operatorname{sech}^2 t-1=-\tanh^2 t\), so

\(\frac{\mathrm{d}x}{\mathrm{d}t}=-\tanh^3 t\).

Hence

\(\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\frac{\mathrm{d}y}{\mathrm{d}t}}{\frac{\mathrm{d}x}{\mathrm{d}t}}=\frac{4\tanh^3 t\,\operatorname{sech}^2 t}{-\tanh^3 t}=-4\operatorname{sech}^2 t\).

(c) From part (b),

\(\frac{\mathrm{d}y}{\mathrm{d}x}=-4\operatorname{sech}^2 t\).

Differentiate with respect to \(t\):

\(\frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)=-4\frac{\mathrm{d}}{\mathrm{d}t}(\operatorname{sech}^2 t)=8\operatorname{sech}^2 t\tanh t\).

Therefore

\(\frac{\mathrm{d}^2y}{\mathrm{d}x^2}=\frac{\frac{\mathrm{d}}{\mathrm{d}t}(\frac{\mathrm{d}y}{\mathrm{d}x})}{\frac{\mathrm{d}x}{\mathrm{d}t}}=\frac{8\operatorname{sech}^2 t\tanh t}{-\tanh^3 t}=-8\frac{\operatorname{sech}^2 t}{\tanh^2 t}\).

We are told that

\(-8\frac{\operatorname{sech}^2 t}{\tanh^2 t}=-\frac92\).

So

\(8\operatorname{sech}^2 t=\frac92\tanh^2 t\).

Using \(\operatorname{sech}^2 t=1-\tanh^2 t\),

\(8(1-\tanh^2 t)=\frac92\tanh^2 t\).

Multiply by 2:

\(16-16\tanh^2 t=9\tanh^2 t\), so \(16=25\tanh^2 t\).

Hence \(\tanh^2 t=\frac{16}{25}\). Since \(t\gt 0\), \(\tanh t=\frac45\).

Then

\(\operatorname{sech}^2 t=1-\frac{16}{25}=\frac{9}{25}\), so \(\operatorname{sech} t=\frac35\).

Now

\(x=\frac12\tanh^2 t+\ln(\operatorname{sech} t)=\frac12\cdot\frac{16}{25}+\ln\frac35=\frac{8}{25}+\ln\frac35\),

\(y=1+\tanh^4 t=1+\left(\frac{16}{25}\right)^2=1+\frac{256}{625}=\frac{881}{625}\).

So the required point is \(\left(\frac{8}{25}+\ln\frac35,\;\frac{881}{625}\right)\).