Solution: Note that \begin{align*} f(x) = x - 1 & \implies f(f(x)) = (x - 1) - 1 = x - 2 \\ & \implies g(f(f(x))) = e^{x-2}. \end{align*} Therefore, the given differential equation reduces to \begin{equation}\label{eq:23May2025-1} \frac{\mathrm{d}y}{\mathrm{d}x} = e^{-2 \sqrt{x} } \cdot e^{x-2} - \frac{y}{\sqrt{x} } \implies \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{y}{\sqrt{x} } = e^{-2\sqrt{x} + x - 2}. \end{equation} This is a linear ODE which can be solved by finding the integrating factor. The integrating factor is given by \begin{align*} \mu = e^{\int \frac{1}{\sqrt{x} }\mathrm{d} x} & = e^{2\sqrt{x} }. \end{align*}

Multiplying the integrating factor to the differential equation \eqref{eq:23May2025-1}, we get \begin{align*} e^{2\sqrt{x} } \left( \frac{\mathrm{d}y}{\mathrm{d}x} + \frac{y}{\sqrt{x} } \right) = e^{x-2} \\ \implies & d\left( e^{2\sqrt{x} }\cdot y \right) = e^{x-2} \\ \implies & e^{2\sqrt{x} } \cdot y = \int e^{x-2} + c \\ \implies & e^{2\sqrt{x} }\cdot y = e^{x-2} + c \\ \implies & y(x) = e^{x - 2 - 2\sqrt{x} } + c e^{-2\sqrt{x} }. \end{align*} Given that $y(0) = 0$, gives, $c = -e^{-2} $. Thus, \begin{align*} y(x) = e^{x-2 - 2\sqrt{x} } - e^{-2\sqrt{x} - 2}. \end{align*} So, \begin{align*} y(1) = e^{-3} - e^{-4} = \frac{e - 1}{e^4}. \end{align*}