The differential form of the equation $(y-a{)}^2=2bx$ is:

Differentiating $(y-a{)}^2=2bx$ with respect to $x$:
$$2(y-a)\frac{dy}{dx}=2b⟹(y-a)\frac{dy}{dx}=b\dots (1)$$
Differentiating again:
$$(y-a)\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=0\dots (2)$$
From the original equation $(y-a{)}^2=2bx$, we have $y-a=\pm \sqrt{2bx}$.From (1), $(y-a)=\frac{b}{dy/dx}$.Substitute this into (2):
$$\frac{b}{dy/dx}\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^2=0$$
$$b\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^3=0$$
Also from (1) and original equation:$b^2/(dy/dx{)}^2=2bx⟹b=2x(dy/dx{)}^2$.Substitute $b$ back:
$$2x{\left(\frac{dy}{dx}\right)}^2\frac{d^{2}y}{d{x}^2}+{\left(\frac{dy}{dx}\right)}^3=0$$
Divide by $(dy/dx{)}^2$:
$$2x\frac{d^{2}y}{d{x}^2}+\frac{dy}{dx}=0$$