Let $0<a<1$. Find all functions $f: \mathbb{R} \rightarrow \mathbb{R}$ continuous at $x = 0$ such that $f(x) + f(ax) = x,\, \forall x \in \mathbb{R}$
1. **Given the functional equation:**
   \[
   f(x) + f(ax) = x \quad \forall x \in \mathbb{R}
   \]
   where \(0 < a < 1\), and \(f\) is continuous at \(x = 0\).

2. **Assume a linear solution of the form \(f(x) = cx\):**
   \[
   f(x) = cx
   \]
   Substituting into the functional equation:
   \[
   cx + c(ax) = x
   \]
   Simplifying, we get:
   \[
   cx + acx = x
   \]
   \[
   c(1 + a)x = x
   \]
   For this to hold for all \(x\), we must have:
   \[
   c(1 + a) = 1
   \]
   Solving for \(c\):
   \[
   c = \frac{1}{1 + a}
   \]
   Therefore, one potential solution is:
   \[
   f(x) = \frac{x}{1 + a}
   \]

3. **To check if there are any other solutions, consider the reduced equation:**
   \[
   f(x) + f(ax) = 0
   \]
   Let \(g(x) = f(x) - \frac{x}{1 + a}\). Then:
   \[
   g(x) + g(ax) = f(x) - \frac{x}{1 + a} + f(ax) - \frac{ax}{1 + a}
   \]
   Using the original functional equation \(f(x) + f(ax) = x\), we get:
   \[
   g(x) + g(ax) = x - \frac{x}{1 + a} - \frac{ax}{1 + a}
   \]
   Simplifying the right-hand side:
   \[
   x - \frac{x}{1 + a} - \frac{ax}{1 + a} = x - \frac{x(1 + a)}{1 + a} = x - x = 0
   \]
   Thus:
   \[
   g(x) + g(ax) = 0
   \]

4. **Analyzing the reduced equation \(g(x) + g(ax) = 0\):**
   \[
   g(x) = -g(ax)
   \]
   Applying this iteratively:
   \[
   g(x) = -g(ax) = g(a^2x) = -g(a^3x) = \cdots
   \]
   Since \(0 < a < 1\), as \(n \to \infty\), \(a^n \to 0\). By the continuity of \(f\) at \(x = 0\), we have:
   \[
   g(0) = 0
   \]
   Therefore, \(g(x) = 0\) for all \(x\), implying:
   \[
   f(x) = \frac{x}{1 + a}
   \]

5. **Conclusion:**
   The only continuous function \(f: \mathbb{R} \to \mathbb{R}\) at \(x = 0\) that satisfies the given functional equation is:
   \[
   f(x) = \frac{x}{1 + a}
   \]

The final answer is \(\boxed{f(x) = \frac{x}{1 + a}}\).