Find all continuous functions $f : \mathbb{R} \to \mathbb{R}$ such that$$f(-x)+\int \limits_0^xtf(x-t)dt=x,\ \forall\ x\in \mathbb{R}$$
1. **Given Equation:**
   \[
   f(-x) + \int_0^x t f(x-t) \, dt = x, \quad \forall x \in \mathbb{R}
   \]

2. **Rewriting the Integral:**
   \[
   f(-x) = x - \int_0^x t f(x-t) \, dt \quad \text{(1)}
   \]

3. **Changing the Variable in the Integral:**
   Let \( u = x - t \). Then \( du = -dt \) and when \( t = 0 \), \( u = x \), and when \( t = x \), \( u = 0 \). Thus,
   \[
   \int_0^x t f(x-t) \, dt = \int_x^0 (x-u) f(u) \, (-du) = \int_0^x (x-u) f(u) \, du
   \]

4. **Simplifying the Integral:**
   \[
   \int_0^x (x-u) f(u) \, du = x \int_0^x f(u) \, du - \int_0^x u f(u) \, du
   \]

5. **Substituting Back into the Equation:**
   \[
   f(-x) = x - \left( x \int_0^x f(u) \, du - \int_0^x u f(u) \, du \right)
   \]
   \[
   f(-x) = x - x \int_0^x f(u) \, du + \int_0^x u f(u) \, du
   \]

6. **Differentiating Both Sides with Respect to \( x \):**
   \[
   \frac{d}{dx} f(-x) = -f'(-x)
   \]
   \[
   \frac{d}{dx} \left( x - x \int_0^x f(u) \, du + \int_0^x u f(u) \, du \right)
   \]
   Using the Leibniz rule for differentiation under the integral sign:
   \[
   \frac{d}{dx} \left( x \int_0^x f(u) \, du \right) = \int_0^x f(u) \, du + x f(x)
   \]
   \[
   \frac{d}{dx} \left( \int_0^x u f(u) \, du \right) = x f(x)
   \]
   Therefore,
   \[
   -f'(-x) = 1 - \int_0^x f(u) \, du \quad \text{(2)}
   \]

7. **Differentiating Equation (2) Again:**
   \[
   \frac{d}{dx} \left( -f'(-x) \right) = -f''(-x)
   \]
   \[
   \frac{d}{dx} \left( 1 - \int_0^x f(u) \, du \right) = -f(x)
   \]
   Thus,
   \[
   f''(-x) = -f(x) \quad \text{(3)}
   \]

8. **Differentiating Equation (3) Twice:**
   \[
   \frac{d^2}{dx^2} f''(-x) = f^{(4)}(-x)
   \]
   \[
   \frac{d^2}{dx^2} (-f(x)) = -f''(x)
   \]
   Therefore,
   \[
   f^{(4)}(-x) = -f''(x) \quad \text{(4)}
   \]

9. **Changing \( x \) to \( -x \) in Equation (4):**
   \[
   f^{(4)}(x) = f(x)
   \]
   This is a linear differential equation with the general solution:
   \[
   f(x) = c_1 e^x + c_2 e^{-x} + c_3 \cos x + c_4 \sin x \quad \text{(5)}
   \]

10. **Using Initial Conditions:**
    From Equation (3) and (5):
    \[
    f''(-x) = -f(x) \implies c_1 = -c_2, \quad c_4 = 0
    \]
    Since \( f(0) = 0 \):
    \[
    c_3 = 0
    \]
    Since \( f'(0) = -1 \):
    \[
    c_1 - c_2 = -1 \implies c_1 = -\frac{1}{2}, \quad c_2 = \frac{1}{2}
    \]
    Therefore,
    \[
    f(x) = -\frac{1}{2} e^x + \frac{1}{2} e^{-x} = -\sinh(x)
    \]

11. **Verification:**
    Substitute \( f(x) = -\sinh(x) \) back into the original equation to verify it satisfies the given condition.

The final answer is \( \boxed{ f(x) = -\sinh(x) } \).