Find the coordinates of the points of intersection of the curve $x + 3 = 2xy$ and the curve $y = \frac{2}{\sqrt{x}}$.

1. Substitute the expression for $y$ from the second equation into the first equation.

Substitute $y = \frac{2}{\sqrt{x}}$ into $x + 3 = 2xy$:
$x + 3 = 2x\left(\frac{2}{\sqrt{x}}\right)$

2. Simplify the equation to form a quadratic equation in terms of $\sqrt{x}$.

$x + 3 = 4\sqrt{x}$
$x - 4\sqrt{x} + 3 = 0$

3. Solve the quadratic equation for $\sqrt{x}$.

Let $u = \sqrt{x}$. Then $u^2 = x$.
$u^2 - 4u + 3 = 0$
$(u - 1)(u - 3) = 0$
$u = 1 \quad \text{or} \quad u = 3$
$\sqrt{x} = 1 \quad \text{or} \quad \sqrt{x} = 3$

4. Find the values of $x$.

$x = 1^2 = 1 \quad \text{or} \quad x = 3^2 = 9$

5. Find the corresponding values of $y$ and state the coordinates.

When $x = 1$, $y = \frac{2}{\sqrt{1}} = 2$.
When $x = 9$, $y = \frac{2}{\sqrt{9}} = \frac{2}{3}$.

The coordinates of the points of intersection are $(1, 2)$ and $\left(9, \frac{2}{3}\right)$.