$\frac{d^{2}x}{dy^{2}}$ equals

d²x/dy² equals $y=y(x) \iff x=x(y) \ (\text{inverse}), \quad \text{find } \dfrac{d^{2}x}{dy^{2}}$ $\text{in terms of } \dfrac{dy}{dx} \text{ and } \dfrac{d^{2}y}{dx^{2}}$ $(1)\ (y'')^{-1}, \quad (2)\ -(y'')^{-1}(y')^{-3}$ $(3)\ (y'')(y')^{-2}, \quad (4)\ -(y'')(y')^{-3}$ $1.\ \dfrac{dx}{dy}=\left(\dfrac{dy}{dx}\right)^{-1}$ $2.\ \dfrac{d}{dy}=\dfrac{dx}{dy}\cdot\dfrac{d}{dx}$ $3.\ \dfrac{d^{2}x}{dy^{2}} \Rightarrow \text{match the option}$ $\dfrac{dx}{dy}=\dfrac{1}{dy/dx}=(y')^{-1}$ $\dfrac{d}{dy}=\dfrac{dx}{dy}\cdot\dfrac{d}{dx}=(y')^{-1}\dfrac{d}{dx}$ $\dfrac{d^{2}x}{dy^{2}}=\dfrac{d}{dy}\left[(y')^{-1}\right]$ $=(y')^{-1}\dfrac{d}{dx}\left[(y')^{-1}\right]$ $\dfrac{d}{dx}\left[(y')^{-1}\right]=-1\cdot(y')^{-2}\cdot y''=-(y')^{-2}y''$ $\dfrac{d^{2}x}{dy^{2}}=(y')^{-1}\cdot\left(-(y')^{-2}y''\right)$ $=-(y')^{-3}\,y''=-\left(\dfrac{d^{2}y}{dx^{2}}\right)\left(\dfrac{dy}{dx}\right)^{-3}$ $y=x^{2}:\ y'=2x,\ y''=2$ $x=\sqrt{y}:\ \dfrac{d^{2}x}{dy^{2}}=-\dfrac{1}{4y^{3/2}}=-\dfrac{2}{(2x)^{3}} \ \checkmark$ $\boxed{\text{option }(4)}$