Find all real-valued continuous functions defined on the interval $(-1, 1)$ such that $(1-x^{2}) f(\frac{2x}{1+x^{2}}) = (1+x^{2})^{2}f(x)$ for all $x$.
1. Define the function $\varphi: (-1,1) \to \mathbb{R}$ by $\varphi(x) = \frac{2x}{1+x^2}$. We need to analyze the properties of $\varphi$.
   
2. Compute the derivative of $\varphi$:
   \[
   \varphi'(x) = \frac{d}{dx} \left( \frac{2x}{1+x^2} \right) = \frac{2(1+x^2) - 2x(2x)}{(1+x^2)^2} = \frac{2(1 - x^2)}{(1 + x^2)^2}
   \]
   Since $\varphi'(x) > 0$ for all $x \in (-1,1)$, $\varphi$ is strictly increasing on $(-1,1)$.

3. Evaluate the limits of $\varphi(x)$ as $x$ approaches the boundaries of the interval:
   \[
   \lim_{x \to -1} \varphi(x) = -1 \quad \text{and} \quad \lim_{x \to 1} \varphi(x) = 1
   \]
   This shows that $\varphi$ maps $(-1,1)$ onto $(-1,1)$.

4. Analyze the behavior of $\varphi(x)$ compared to $x$:
   \[
   \varphi(x) - x = \frac{2x}{1+x^2} - x = \frac{2x - x(1+x^2)}{1+x^2} = \frac{x(1-x^2)}{1+x^2}
   \]
   - For $x \in (0,1)$, $\varphi(x) > x$.
   - For $x \in (-1,0)$, $\varphi(x) < x$.

5. Simplify the given functional equation:
   \[
   (1-x^2) f\left( \frac{2x}{1+x^2} \right) = (1+x^2)^2 f(x)
   \]
   Substitute $y = \varphi(x) = \frac{2x}{1+x^2}$, then $x = \varphi^{-1}(y)$ and we have:
   \[
   1 - y^2 = \frac{(1-x^2)^2}{(1+x^2)^2}
   \]
   Therefore, the equation becomes:
   \[
   (1 - y^2) f(y) = (1 - x^2) f(x)
   \]

6. Consider the behavior of $\varphi^{-n}(x)$ as $n \to \infty$:
   \[
   \varphi^{-n}(x) = \underbrace{\varphi^{-1}(\cdots \varphi^{-1}(x) \cdots)}_{n \text{ times}} \to 0 \quad \text{for all } x \in (-1,1)
   \]
   Thus, we have:
   \[
   (1 - x^2) f(x) = (1 - (\varphi^{-n}(x))^2) f(\varphi^{-n}(x)) \to f(0)
   \]

7. Since $f(0)$ is a constant, let $f(0) = c$. Then:
   \[
   (1 - x^2) f(x) = c \implies f(x) = \frac{c}{1 - x^2}
   \]

The final answer is $\boxed{f(x) = \frac{c}{1 - x^2}}$ for all $c \in \mathbb{R}$.