Let's find all functions $f: (0, \infty) \rightarrow (0, \infty)$ that satisfy the functional equation:

$$xf(x^2)f(f(y)) + f(yf(x)) = f(xy) \left( f(f(x^2)) + f(f(y^2)) \right).$$

To solve this problem, consider the possibility $f(x) = \frac{1}{x}$. We will verify if this satisfies the given functional equation for all $x, y \in (0, \infty)$.

Verification:

Suppose $f(x) = \frac{1}{x}$.

1. Compute each term in the equation with this $f(x)$:

$$f(x^2) = \frac{1}{x^2}, \quad f(f(y)) = \frac{1}{\frac{1}{y}} = y, \quad f(yf(x)) = f\left(\frac{y}{x}\right) = \frac{x}{y}.$$

$$f(xy) = \frac{1}{xy}, \quad f(f(x^2)) = \frac{1}{\frac{1}{x^2}} = x^2, \quad f(f(y^2)) = \frac{1}{\frac{1}{y^2}} = y^2.$$

2. Substitute these into the given equation:

- Left-hand side:
$$x \cdot \frac{1}{x^2} \cdot y + \frac{x}{y} = \frac{y}{x} + \frac{x}{y}.$$

- Right-hand side:
$$\frac{1}{xy} \cdot (x^2 + y^2) = \frac{x^2 + y^2}{xy}.$$

3. Check if these expressions are equal:

- Simplify the left-hand side:
$$\frac{y}{x} + \frac{x}{y} = \frac{y^2 + x^2}{xy}.$$

- Simplified right-hand side is:
$$\frac{x^2 + y^2}{xy}.$$

Both simplify to the same expression $\frac{x^2 + y^2}{xy}$, hence $f(x) = \frac{1}{x}$ satisfies the functional equation.

Thus, the only function that satisfies the given equation is:

$$\boxed{f(x) = \frac{1}{x}}$$