Equation of the ellipse whose axes are the axes of coordinates and which passes through the point $$(-3, 1)$$ and has eccentricity $$\sqrt{\dfrac{2}{5}}$$ is:

The centre of the required ellipse is the origin and its axes coincide with the coordinate axes, so its equation can be written in the standard form

$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,$$
where $$a$$ is the semi-major axis, $$b$$ is the semi-minor axis and, by definition, $$a \gt b\;.$$

The eccentricity of such an ellipse is defined as

$$e=\sqrt{1-\frac{b^{2}}{a^{2}}}\;.$$

It is given that $$e=\sqrt{\frac{2}{5}}\;,$$ hence

$$e^{2}=1-\frac{b^{2}}{a^{2}}=\frac{2}{5}\quad\Longrightarrow\quad \frac{b^{2}}{a^{2}}=1-\frac{2}{5}=\frac{3}{5}.$$

Therefore

$$b^{2}=\frac{3}{5}\,a^{2}\;.\qquad -(1)$$

The ellipse passes through the point $$(-3,1)\,,$$ so

$$\frac{(-3)^{2}}{a^{2}}+\frac{(1)^{2}}{b^{2}}=1 \;\Longrightarrow\; \frac{9}{a^{2}}+\frac{1}{b^{2}}=1\;.\qquad -(2)$$

Substituting $$b^{2}=\dfrac{3a^{2}}{5}$$ from (1) into (2):

$$\frac{9}{a^{2}}+\frac{1}{\,\dfrac{3a^{2}}{5}\,}=1 \;\Longrightarrow\; \frac{9}{a^{2}}+\frac{5}{3a^{2}}=1 \;\Longrightarrow\; \frac{27+5}{3a^{2}}=1 \;\Longrightarrow\; \frac{32}{3a^{2}}=1 \;\Longrightarrow\; a^{2}=\frac{32}{3}\;.$$

Using (1),

$$b^{2}=\frac{3}{5}\cdot\frac{32}{3}=\frac{32}{5}\;.$$

Putting these values of $$a^{2}$$ and $$b^{2}$$ back into the standard equation,

$$\frac{x^{2}}{\,\dfrac{32}{3}\,}+\frac{y^{2}}{\,\dfrac{32}{5}\,}=1.$$

Multiply throughout by the LCM $$32$$:

$$3x^{2}+5y^{2}=32.$$

Re-arranging, the required equation is

$$3x^{2}+5y^{2}-32=0.$$

Hence the correct choice is:
Option D which is: $$3x^{2}+5y^{2}-32=0$$.