1. Form of the ellipse

Since the ellipse has axes along the coordinate axes, its equation must be of the form

$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$

with center at the origin.

2. Use the eccentricity

Given eccentricity

$e = \sqrt{\frac{2}{5}}$

For an ellipse,

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

So,

$\frac{2}{5} = 1 - \frac{b^2}{a^2}$

$\frac{b^2}{a^2} = 1 - \frac{2}{5} = \frac{3}{5}$

Hence,

$b^2 : a^2 = 3:5$

Let

$a^2 = 5k, \qquad b^2 = 3k$

Thus the ellipse becomes

$\frac{x^2}{5k} + \frac{y^2}{3k} = 1$

3. Use the point $(-3,1)$

Since the ellipse passes through $(-3,1)$,

$\frac{(-3)^2}{5k} + \frac{1^2}{3k} = 1$

$\frac{9}{5k} + \frac{1}{3k} = 1$

Taking LCM:

$\frac{27 + 5}{15k} = 1$

$\frac{32}{15k} = 1$

$15k = 32$

$k = \frac{32}{15}$

4. Find the equation

Now,

$a^2 = 5k = 5\cdot \frac{32}{15} = \frac{32}{3}$

$b^2 = 3k = 3\cdot \frac{32}{15} = \frac{32}{5}$

So the equation is

$\frac{x^2}{32/3} + \frac{y^2}{32/5} = 1$

Multiply through by $32$:

$3x^2 + 5y^2 = 32$

or

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

5. Match with the options

This is Option D.

Comparison with stored answer: Stored correct answer is D, which matches our result.