Step 1: Since the centre is at $(0, 0)$ and major axis is on the y-axis, the standard equation of the ellipse is

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

where $a$ is the semi-major axis and $b$ is the semi-minor axis.

Step 2: Since the ellipse passes through point $(3, 2)$, substituting $x = 3$ and $y = 2$:

$$\frac{9}{b^2} + \frac{4}{a^2} = 1 \quad \text{...(i)}$$

Step 3: Since the ellipse passes through point $(1, 6)$, substituting $x = 1$ and $y = 6$:

$$\frac{1}{b^2} + \frac{36}{a^2} = 1 \quad \text{...(ii)}$$

Step 4: Let $\frac{1}{b^2} = u$ and $\frac{1}{a^2} = v$. Then equations (i) and (ii) become:

$$9u + 4v = 1$$ $$u + 36v = 1$$

Step 5: From the second equation, $u = 1 - 36v$. Substituting in the first equation:

$$9(1 - 36v) + 4v = 1$$

which gives

$$9 - 324v + 4v = 1$$

so

$$320v = 8$$

therefore

$$v = \frac{1}{40}$$

Step 6: Then

$$u = 1 - 36\left(\frac{1}{40}\right) = 1 - \frac{9}{10} = \frac{1}{10}$$

Thus $a^2 = 40$ and $b^2 = 10$.

Step 7: The required equation of the ellipse is

$$\frac{x^2}{10} + \frac{y^2}{40} = 1$$