Find all functions $f: \mathbb{R}^{+} \rightarrow \mathbb{R}$ and $g: \mathbb{R}^{+} \rightarrow \mathbb{R}$ such that
$$f(x^2 + y^2) = g(xy)$$
holds for all $x, y \in \mathbb{R}^{+}$.

1. **Define the functional equation and initial assertion:**
   We are given the functional equation:
   \[
   f(x^2 + y^2) = g(xy)
   \]
   for all \( x, y \in \mathbb{R}^{+} \). Denote this assertion as \( P(x, y) \).

2. **Evaluate the functional equation at specific points:**
   Consider the assertion \( P\left(x, \frac{1}{x}\right) \):
   \[
   f\left(x^2 + \frac{1}{x^2}\right) = g(1)
   \]
   By the AM-GM inequality, we know that:
   \[
   x^2 + \frac{1}{x^2} \ge 2
   \]
   This implies that \( f(x) = g(1) \) for all \( x \ge 2 \).

3. **Use another specific point to show \( g \) is constant:**
   Consider the assertion \( P\left(\frac{x}{2}, 2\right) \):
   \[
   f\left(\left(\frac{x}{2}\right)^2 + 2^2\right) = g\left(\frac{x}{2} \cdot 2\right)
   \]
   Simplifying, we get:
   \[
   f\left(\frac{x^2}{4} + 4\right) = g(x)
   \]
   Since \( \frac{x^2}{4} + 4 \ge 4 \) and we already know \( f(x) = g(1) \) for all \( x \ge 2 \), it follows that:
   \[
   f\left(\frac{x^2}{4} + 4\right) = g(1)
   \]
   Therefore, \( g(x) = g(1) \) for all \( x \in \mathbb{R}^{+} \), which means \( g \) is a constant function.

4. **Conclude that \( f \) is also constant:**
   Since \( g(x) = g(1) \) for all \( x \in \mathbb{R}^{+} \), we have:
   \[
   f(x^2 + y^2) = g(1)
   \]
   for all \( x, y \in \mathbb{R}^{+} \). This implies that \( f \) is also a constant function.

5. **Determine the constant values:**
   Let \( c = g(1) \). Then:
   \[
   f(x) = c \quad \text{and} \quad g(x) = c
   \]
   for all \( x \in \mathbb{R}^{+} \).

The final answer is \( \boxed{ f(x) = c } \) and \( g(x) = c \) for some constant \( c \in \mathbb{R} \).