The differential equation of the family of curves, $$x^2 = 4b(y + b)$$, $$b \in R$$, is.

We are given the one-parameter family of curves

$$x^{2}=4b\,(y+b),\qquad b\in\mathbb R,$$

where $$y=y(x)$$ is the dependent variable and $$b$$ is the parameter. Because only one parameter is present, we expect the required differential equation to be of first order. Our aim is to eliminate $$b$$ by differentiating with respect to $$x$$ and then substituting.

First we differentiate the given relation with respect to $$x$$. Remember the rule $$\dfrac{d}{dx}(x^{n})=nx^{n-1}$$ and that, during differentiation, $$b$$ behaves as a constant:

$$\dfrac{d}{dx}(x^{2})=\dfrac{d}{dx}\!\bigl(4b(y+b)\bigr).$$

On the left we have $$2x.$$ On the right we have $$4b\dfrac{dy}{dx}+0$$ because $$\dfrac{d}{dx}(y)=\dfrac{dy}{dx}=y'$$ and $$\dfrac{d}{dx}(b)=0.$$ Thus

$$2x=4b\,y'.$$

Now we solve this equation for the parameter $$b$$:

$$b=\frac{2x}{4y'}=\frac{x}{2y'}.$$

Next we return to the original family equation and substitute this expression for $$b$$. We have

$$x^{2}=4b\,(y+b)$$

$$\Longrightarrow\;x^{2}=4\left(\frac{x}{2y'}\right)\!\left(y+\frac{x}{2y'}\right).$$

Simplifying step by step, we first combine the constant factors:

$$4\left(\frac{x}{2y'}\right)=\frac{4x}{2y'}=\frac{2x}{y'}.$$

Therefore

$$x^{2}=\frac{2x}{y'}\left(y+\frac{x}{2y'}\right).$$

We now distribute the factor $$\dfrac{2x}{y'}$$ inside the brackets:

$$x^{2}=\frac{2x}{y'}\cdot y\;+\;\frac{2x}{y'}\cdot\frac{x}{2y'}.$$

The first product gives $$\dfrac{2xy}{y'}.$$ For the second product the 2 in the numerator and denominator cancel, giving $$\dfrac{x^{2}}{(y')^{2}}.$$ Hence

$$x^{2}= \frac{2xy}{y'}+\frac{x^{2}}{(y')^{2}}.$$

To remove the denominators we multiply every term by $$(y')^{2}$$ (this is legitimate provided $$y'$$ is finite):

$$x^{2}(y')^{2}=2xy\,y'+x^{2}.$$

Now we rearrange all terms to one side in order to see the structure clearly:

$$x^{2}(y')^{2}-2xy\,y'-x^{2}=0.$$

Because $$x\neq0$$ for a non-trivial curve, we can divide the entire equation by $$x$$ to simplify:

$$x(y')^{2}-2y\,y'-x=0.$$

Finally we isolate the term $$x(y')^{2}$$:

$$x(y')^{2}=x+2y\,y'.$$

This matches exactly the form given in option A:

$$x(y')^{2}=x+2y\,y'.$$

Hence, the correct answer is Option A.