Determine all non-negative integers $n$ for which there exist two relatively prime non-negative integers $x$ and $y$ and a positive integer $k\geqslant 2$ such that $3^n=x^k+y^k$.
1. **Case for \( n \leq 2 \):**
   - For \( n = 0 \), \( 3^0 = 1 \). There are no non-negative integers \( x \) and \( y \) such that \( x^k + y^k = 1 \) for \( k \geq 2 \).
   - For \( n = 1 \), \( 3^1 = 3 \). There are no non-negative integers \( x \) and \( y \) such that \( x^k + y^k = 3 \) for \( k \geq 2 \).
   - For \( n = 2 \), \( 3^2 = 9 \). There are no non-negative integers \( x \) and \( y \) such that \( x^k + y^k = 9 \) for \( k \geq 2 \).

2. **Case for \( n \geq 3 \):**
   - We need to show that there are no solutions for \( n \geq 3 \).
   - Assume \( 3^n = x^k + y^k \) where \( x \) and \( y \) are relatively prime non-negative integers and \( k \geq 2 \).
   - Since \( x \) and \( y \) are relatively prime, \( k \) must be odd. If \( k \) were even, both \( x \) and \( y \) would have to be divisible by 3, contradicting the assumption that they are relatively prime.

3. **Divisibility and LTE Theorem:**
   - Since \( x \) and \( y \) are relatively prime, \( x + y \mid x^k + y^k = 3^n \). Therefore, \( x + y = 3^q \) for some integer \( q \).
   - Using the Lifting The Exponent (LTE) theorem, we have:
     \[
     v_3(x^k + y^k) = v_3(x + y) + v_3(k) = q + v_3(k)
     \]
     where \( v_3 \) denotes the 3-adic valuation.
   - Thus, \( n = q + v_3(k) \). This implies \( v_3(k) = n - q \), and hence \( k \geq 3^{n-q} \).

4. **Bounding \( k \):**
   - Since \( x + y = 3^q \) is odd, without loss of generality, assume \( x > y \).
   - Consider the polynomial expansion:
     \[
     x^k + y^k = (x + y)(x^{k-1} - x^{k-2}y + x^{k-3}y^2 - \cdots - xy^{k-2} + y^{k-1})
     \]
   - If \( y = 1 \), then \( x^k = 3^n - 1 \). There are no solutions for this as \( 3^n - 1 \) is not a perfect \( k \)-th power for \( k \geq 2 \).

5. **Contradiction:**
   - Therefore, \( y \geq 2 \). Then:
     \[
     3^n = x^k + y^k = (x + y)(x^{k-1} - x^{k-2}y + x^{k-3}y^2 - \cdots - xy^{k-2} + y^{k-1}) > 2^{k-1} \cdot 3^q
     \]
   - This implies:
     \[
     3^{n-q} > 2^{k-1}
     \]
   - But we have \( 2^{k-1} \geq k \geq 3^{n-q} \), leading to a contradiction.

6. **Conclusion:**
   - Therefore, the only possible values for \( n \) are \( n = 0, 1, 2 \).

The final answer is \( \boxed{ n = 0, 1, 2 } \).