Given points $(x_1, y_1)$, $(x_2, y_2)$, and $(x_3, y_3)$ lie on a straight line and on the curve $y^2 = x^3$. We need to show that $\frac{x_1}{y_1} + \frac{x_2}{y_2} + \frac{x_3}{y_3} = 0$.

1. Express the points in terms of parameters:
Since the points lie on the curve $y^2 = x^3$, we can write:
$$(x_1, y_1) = (a^2, a^3), \quad (x_2, y_2) = (b^2, b^3), \quad (x_3, y_3) = (c^2, c^3)$$
where $a$, $b$, and $c$ are distinct and non-zero.

2. Collinearity condition:
The points are collinear if the slope between any two pairs of points is the same. Using the points $(a^2, a^3)$, $(b^2, b^3)$, and $(c^2, c^3)$, the collinearity condition can be written as:
$$\frac{b^3 - c^3}{b^2 - c^2} = \frac{a^3 - b^3}{a^2 - b^2}$$

3. Simplify the collinearity condition:
Using the identity for the difference of cubes and squares, we get:
$$\frac{(b - c)(b^2 + bc + c^2)}{(b - c)(b + c)} = \frac{(a - b)(a^2 + ab + b^2)}{(a - b)(a + b)}$$
Simplifying, we obtain:
$$\frac{b^2 + bc + c^2}{b + c} = \frac{a^2 + ab + b^2}{a + b}$$

4. Cross-multiply and expand:
Cross-multiplying both sides, we get:
$$(b^2 + bc + c^2)(a + b) = (a^2 + ab + b^2)(b + c)$$
Expanding both sides:
$$b^2a + b^3 + abc + b^2c + c^2a + c^2b = a^2b + a^2c + ab^2 + abc + b^3 + b^2c$$

5. Simplify the equation:
Canceling common terms on both sides, we get:
$$c^2a + c^2b = a^2b + a^2c$$
Rearranging terms, we have:
$$c^2(a + b) = a^2(b + c)$$

6. Factor and solve:
Factoring out common terms, we get:
$$(a - c)(ab + bc + ca) = 0$$
Since $a$, $b$, and $c$ are distinct, $a \neq c$. Therefore, we must have:
$$ab + bc + ca = 0$$

7. Show the required result:
We need to show that $\frac{x_1}{y_1} + \frac{x_2}{y_2} + \frac{x_3}{y_3} = 0$. Substituting the parameterized points:
$$\frac{x_1}{y_1} + \frac{x_2}{y_2} + \frac{x_3}{y_3} = \frac{a^2}{a^3} + \frac{b^2}{b^3} + \frac{c^2}{c^3} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$$
Since $ab + bc + ca = 0$, we have:
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{ab + bc + ca}{abc} = \frac{0}{abc} = 0$$

Thus, we have shown that $\frac{x_1}{y_1} + \frac{x_2}{y_2} + \frac{x_3}{y_3} = 0$.

$\blacksquare$