To find all functions $f: (0, \infty) \to (0, \infty)$ satisfying the given functional equation:

$$\frac {\left( f(w) \right)^2 + \left( f(x) \right)^2}{f(y^2) + f(z^2) } = \frac {w^2 + x^2}{y^2 + z^2}$$

for all positive real numbers $w, x, y, z$ such that $wx = yz$, we proceed as follows:

### Step 1: Analyze the Functional Equation

The equation provides a relationship between the values of the function at different points. Considering $wx = yz$, the form $wx = yz$ suggests symmetry relationships, which often hints at the possibility of the function being of a simple form, such as a power function.

### Step 2: Assume Specific Forms for $f$

Assume $f(x) = x^k$ for some real number $k$. Substitute into the functional equation:

$$\frac {(w^k)^2 + (x^k)^2}{(y^2)^k + (z^2)^k} = \frac {w^2 + x^2}{y^2 + z^2}$$

Simplify the expressions:

$$\frac {w^{2k} + x^{2k}}{y^{2k} + z^{2k}} = \frac {w^2 + x^2}{y^2 + z^2}$$

This equation should hold for all $w, x, y, z$ such that $wx = yz$.

### Step 3: Analyze Specific Cases

1. **Case 1: $f(x) = x$**

- Substituting $f(x) = x$ gives:
$$\frac {w^2 + x^2}{y^2 + z^2} = \frac {w^2 + x^2}{y^2 + z^2}$$
which holds true.

2. **Case 2: $f(x) = \frac{1}{x}$**

- Substituting $f(x) = \frac{1}{x}$ gives:
$$\frac {\left(\frac{1}{w}\right)^2 + \left(\frac{1}{x}\right)^2}{\left(\frac{1}{y^2}\right) + \left(\frac{1}{z^2}\right)} = \frac {w^2 + x^2}{y^2 + z^2}$$
Simplifying the left side:
$$\frac{\frac{1}{w^2} + \frac{1}{x^2}}{\frac{y^2 + z^2}{y^2z^2}} = \frac{w^2 + x^2}{y^2 + z^2}$$
Cross-multiplying confirms equality, as the terms match.

Thus, both functions satisfy the given equation. Therefore, the solutions to the functional equation are:

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

These are the only functions from the positive reals to the positive reals satisfying the given condition.