Solution: The given equation is \begin{equation}\label{eq:11Jan2025-1} x^4 + y^4 = 1. \end{equation} Differentiating equation \eqref{eq:11Jan2025-1}, we get \begin{equation}\label{eq:11Jan2025-2} 4x^3 + 4 y^3 \frac{\mathrm{d}y}{\mathrm{d}x} = 0. \end{equation} Again differentiating the equation \eqref{eq:11Jan2025-2} with respect to $x$, we get \begin{equation}\label{eq:11Jan2025-3} 12x^2 + 12 y^2 \frac{\mathrm{d}y}{\mathrm{d}x} + 4y^3 \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0 . \end{equation} Using equation \eqref{eq:11Jan2025-2}, we have \begin{equation}\label{eq:11Jan2025-4} \frac{\mathrm{d}y}{\mathrm{d}x} = - \frac{x^3}{y^3}. \end{equation} Substituting the above into equation \eqref{eq:11Jan2025-3}, we get \begin{align*} & 12x^2 + 12y^2 \left[ -\frac{x^3}{y^3} \right] + 4y^3 \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = 0 \\ \implies & 4y^3 \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = -12x^2 + \frac{12x^3}{y} \\ \implies & \frac{\mathrm{d}^2y}{\mathrm{d}x^2} = \frac{3x^3}{y^4} - \frac{3x^2}{y^3} = \frac{3x^3 - 3x^2 y}{y^4}. \end{align*} Thus, \[ \textcolor{blue}{\boxed{ \frac{\mathrm{d}^2 y}{\mathrm{d} x^2} = \frac{3x^3 - 3x^2 y}{y^4} }} \]