$$\begin{aligned} \& R O C=\text { Radius of curvature at point } A \\ \& \text { Curvature }=\frac{1}{R O C}=\frac{\left|\frac{d^2 y}{d x^2}\right|}{\left(1+\left(\frac{d y}{d x}\right)^2\right)^{\frac{3}{2}}}=\frac{\left|\frac{d^2 y}{d x^2}\right|}{(1+0)^{\frac{3}{2}}}=\frac{d^2 y}{d x^2} \quad\left[\because \frac{d y}{d x}=\tan \theta=0\right] \\ \& \Delta P=S \times \text { curvature } \\ \& \Rightarrow \rho g y=S \frac{d^2 y}{d x^2} \\ \& \therefore \frac{d^2 y}{d x^2}=\frac{\rho g y}{S} \end{aligned}$$

Alternate Solution :

For the given element, (consider length d in Z direction) Net force in upward direction $=$ Weight $(\mathrm{S} \sin (\theta+\mathrm{d} \theta)-\mathrm{S} \sin \theta) \mathrm{d}=\mathrm{mg}$

Angle is small $\therefore \sin \theta \approx \theta$

$$ \begin{aligned} \& \Rightarrow \frac{d \theta}{y d x}=\frac{\rho g}{S} \\ \& \tan \theta=\frac{d y}{d x} \Rightarrow \text { Differentiating wrt } x \\ \& \sec ^2 \theta \frac{d \theta}{d x}=\frac{d^2 y}{d x^2} \end{aligned} $$

Put $d \theta$ from (2) in (1) $d$ take $\cos \theta \approx 1$, we get

$$ \frac{d^2 y}{d x^2}=\frac{\rho g y}{S} $$