If $y$ is a function of $x$, then $\frac{d^{2}y}{d{x}^2}+y·\frac{dy}{dx}=0$. If $x$ is a function of $y$, then the equation becomes

The given equation is $\frac{d^{2}y}{d{x}^2}+y·\frac{dy}{dx}=0$

Here, $\frac{dy}{dx}=\frac{1}{{\displaystyle \frac{dx}{dy}}}$

Substituting $\frac{dy}{dx}=\frac{1}{{\displaystyle \frac{dx}{dy}}}$in the given equation.

$$\begin{gathered} \frac{d^{2}y}{d{x}^2}=\frac{d}{dx}\left(\frac{1}{{\displaystyle \frac{dx}{dy}}}\right) \\ =\frac{d}{dy}\left(\frac{1}{{\displaystyle \frac{dx}{dy}}}\right)·\frac{dy}{dx} \\ =-\frac{1}{{\left({\displaystyle \frac{dx}{dy}}\right)}^2}·\frac{d^{2}x}{d{y}^2}\left(\frac{1}{{\displaystyle \frac{dx}{dy}}}\right) \\ \frac{d^{2}y}{d{x}^2}=-\frac{\frac{d^{2}x}{d{y}^2}}{{\left(\frac{dx}{dy}\right)}^3} \end{gathered}$$

Now, $\frac{d^{2}y}{d{x}^2}+y·\frac{dy}{dx}=0$

$$\begin{gathered} -\frac{\frac{d^{2}x}{d{y}^2}}{{\left(\frac{dx}{dy}\right)}^3}+y\frac{dy}{dx}=0 \\ \frac{d^{2}x}{d{y}^2}-y\frac{dy}{dx}{\left(\frac{dx}{dy}\right)}^3=0 \\ \frac{d^{2}x}{d{y}^2}-y{\left(\frac{dx}{dy}\right)}^2=0 \end{gathered}$$

For the given equation, if $x$ is a function of $y$, then the equation becomes $\frac{d^{2}x}{d{y}^2}-y{\left(\frac{dx}{dy}\right)}^2=0$.

Therefore, the correct answer is option $\left(3\right)$.