If $\mathrm{y}={\mathrm{e}}^{\mathrm{nx}},$ then $\left(\frac{{\mathrm{d}}^2\mathrm{y}}{{\mathrm{dx}}^2}\right)\left(\frac{{\mathrm{d}}^2\mathrm{x}}{{\mathrm{dy}}^2}\right)$ is equal to

The given function is $y=e^{nx}$

Differentiate both sides 

$\frac{dy}{dx}=n{e}^{nx}$

Again differentiate both sides 

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=n^2{e}^{nx} \\ now y=e^{nx}\Rightarrow nx=\log y \\ \Rightarrow n\frac{dx}{dy}=\frac{1}{y} \\ \Rightarrow n\frac{d{ }^{2}x}{d{y}^2}=\frac{-1}{y^2}\Rightarrow \frac{d{ }^{2}x}{d{y}^2}=\frac{-1}{n{y}^2} \\ \Rightarrow \frac{d^{2}y}{d{x}^2}\frac{d{ }^{2}x}{d{y}^2}=n^{2}y\times \frac{-1}{n{y}^2}=-\frac{n}{y}=-n{e}^{-nx} \end{gathered}$$