If $\frac{x^2}{a^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$, then $\frac{\mathrm{d}^2 \mathrm{y}}{\mathrm{d} x^2}$ is

Implicit differentiation of the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ yields:

$\frac{2x}{a^2} + \frac{2y}{b^2} \frac{dy}{dx} = 0$

Solving for the first derivative gives $\frac{dy}{dx} = -\frac{b^2x}{a^2y}$.

Differentiating again using the quotient rule:
$\frac{d^2y}{dx^2} = -\frac{1}{a^2} \left(\frac{y \cdot b^2 - b^2x \cdot \frac{dy}{dx}}{y^2}\right)$

Substituting the expression for $\frac{dy}{dx}$ produces:
$\frac{d^2y}{dx^2} = -\frac{1}{a^2} \left(\frac{b^2y + \frac{b^4x^2}{a^2y}}{y^2}\right) = -\frac{b^2(a^2y^2 + b^2x^2)}{a^4y^3}$

The original equation implies $b^2x^2 + a^2y^2 = a^2b^2$, which simplifies the expression to:

$\frac{d^2y}{dx^2} = -\frac{b^4}{a^2y^3}$

This corresponds to option D.