The differential equation representing the family of ellipses having foci either on the x-axis or on the y-axis, center at the origin and passing through the point (0, 3) is:

We begin by recalling the standard equation of an ellipse whose centre is at the origin. If the major axis is along the x-axis, the equation is written as $$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1,\qquad a\gt b\gt 0.$$ In this form the two foci lie on the x-axis at the points $$(\pm c,0),\qquad\text{where }c^{2}=a^{2}-b^{2}.$$ If instead the major axis is along the y-axis, the equation is $$\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1,\qquad a\gt b\gt 0,$$ and the foci are on the y-axis.

The given family of curves must always pass through the fixed point $$(0,3).$$ We substitute this point into both possible standard forms and see what restriction is produced.

First take the form with foci on the x-axis:

$$\dfrac{0^{2}}{a^{2}}+\dfrac{3^{2}}{b^{2}}=1\;\;\Longrightarrow\;\;\dfrac{9}{b^{2}}=1\;\;\Longrightarrow\;\;b^{2}=9.$$

Thus, for every such ellipse that goes through $$(0,3)$$, the semi-minor axis is fixed at $$b=3$$ while the semi-major axis $$a$$ can vary with the only condition $$a\gt 3$$ (so that the curve is genuinely an ellipse, not a circle). Hence the whole family is described by the single-parameter equation

$$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{9}=1,\qquad a\gt 3. \quad -(1)$$

Now look at the alternative form with foci on the y-axis:

$$\dfrac{0^{2}}{b^{2}}+\dfrac{3^{2}}{a^{2}}=1\;\;\Longrightarrow\;\;\dfrac{9}{a^{2}}=1\;\;\Longrightarrow\;\;a^{2}=9.$$

But if $$a^{2}=9$$ we would have $$a=b=3$$, which degenerates to a circle whose foci coincide at the origin; such a curve does not possess two distinct foci on the y-axis, so it is excluded by the wording of the problem. Hence the only admissible family is the one given in (1). The single unknown parameter is $$a$$, so we expect a first-order differential equation.

For convenience we rename $$a^{2}=A,\qquad A\gt 9,$$ so that equation (1) becomes

$$\dfrac{x^{2}}{A}+\dfrac{y^{2}}{9}=1. \quad -(2)$$

We now eliminate the parameter $$A$$. Differentiating (2) implicitly with respect to $$x$$, we use the usual differentiation rules: $$\dfrac{d}{dx}\bigl(x^{2}\bigr)=2x,\qquad \dfrac{d}{dx}\bigl(y^{2}\bigr)=2y\,y'.$$ Performing the differentiation gives

$$\dfrac{2x}{A}+\dfrac{2y}{9}\,y'=0. \quad -(3)$$

From the original equation (2) we solve for the term $$\dfrac{x^{2}}{A}$$ so that we can replace $$\dfrac{1}{A}$$ later. Re-arranging (2) yields

$$\dfrac{x^{2}}{A}=1-\dfrac{y^{2}}{9}. \quad -(4)$$

Dividing (4) by $$x$$ gives an explicit expression for $$\dfrac{x}{A}$$:

$$\dfrac{x}{A}=\dfrac{1-\dfrac{y^{2}}{9}}{x}. \quad -(5)$$

We now return to (3) and divide both sides by $$2$$ to simplify:

$$\dfrac{x}{A}+\dfrac{y}{9}\,y'=0. \quad -(6)$$

Substituting the value of $$\dfrac{x}{A}$$ from (5) into (6) gives

$$\dfrac{1-\dfrac{y^{2}}{9}}{x}+\dfrac{y}{9}\,y'=0. \quad -(7)$$

To clear denominators we multiply the entire equation (7) by $$9x$$, obtaining

$$9\Bigl(1-\dfrac{y^{2}}{9}\Bigr)+x\,y\,y'=0.$$

Expanding the bracket and simplifying we see

$$9-y^{2}+x\,y\,y'=0. \quad -(8)$$

Finally we arrange the terms in a more conventional order:

$$x\,y\,y'-y^{2}+9=0. \quad -(9)$$

The equation (9) is a first-order differential equation containing no arbitrary constants; therefore it represents the required family of ellipses. Comparing (9) with the options supplied, we see that it matches option A exactly.

Hence, the correct answer is Option A.