Step 1: Standard Form Transformation: Divide the entire equation $x^2 + 2y^2 = 3$ by 3 to set the right side to 1:
$$\frac{x^2}{3} + \frac{2y^2}{3} = 1 \implies \frac{x^2}{3} + \frac{y^2}{3/2} = 1$$
Step 2: Identify Semi-axes: Comparing with $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$:
$a^2 = 3$ and $b^2 = \frac{3}{2}$.
Since $a^2 \gt b^2$, this is a horizontal ellipse where $a$ is the semi-major axis and $b$ is the semi-minor axis.
Step 3: Calculate Eccentricity ($e$): The formula for eccentricity $e$ is:
$$e = \sqrt{1 - \frac{b^2}{a^2}}$$
Substituting the values:
$$e = \sqrt{1 - \frac{3/2}{3}} = \sqrt{1 - \frac{1}{2}}$$
$$e = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}$$
Thus, the eccentricity of the given ellipse is $1/\sqrt{2}$.