Find all functions $ f\in\mathcal{C}^1 [0,1] $ that satisfy $ f(1)=-1/6 $ and
$$ \int_0^1 \left( f'(x) \right)^2 dx\le 2\int_0^1 f(x)dx. $$
1. **Define the function \( g(x) \):**
   Let \( g(x) = f(1-x) \). This transformation is useful because it simplifies the boundary condition to \( g(0) = f(1) = -\frac{1}{6} \).

2. **Integral transformation:**
   The integral inequality remains unchanged under this transformation:
   \[
   \int_0^1 (g'(x))^2 \, dx \leq 2 \int_0^1 g(x) \, dx.
   \]

3. **Express \( b \) in terms of \( g \):**
   Let \( a = g(0) = -\frac{1}{6} \) and \( b = \int_0^1 g(x) \, dx \). We can express \( b \) as:
   \[
   b = \int_0^1 g(x) \, dx = \int_0^1 \left( a + \int_0^x g'(t) \, dt \right) dx.
   \]
   By changing the order of integration, we get:
   \[
   b = a + \int_0^1 \int_0^x g'(t) \, dt \, dx = a + \int_0^1 \int_t^1 g'(t) \, dx \, dt = a + \int_0^1 (1-t) g'(t) \, dt.
   \]

4. **Apply the Cauchy-Schwarz inequality:**
   Using the Cauchy-Schwarz inequality, we have:
   \[
   (b - a)^2 \leq \left( \int_0^1 (1-t)^2 \, dt \right) \left( \int_0^1 (g'(t))^2 \, dt \right).
   \]
   Calculate the integral:
   \[
   \int_0^1 (1-t)^2 \, dt = \int_0^1 (1 - 2t + t^2) \, dt = \left[ t - t^2 + \frac{t^3}{3} \right]_0^1 = 1 - \frac{1}{2} + \frac{1}{3} = \frac{1}{3}.
   \]
   Therefore:
   \[
   (b + \frac{1}{6})^2 \leq \frac{1}{3} \int_0^1 (g'(t))^2 \, dt = \frac{c}{3}.
   \]

5. **Given condition \( c \leq 2b \):**
   We are given \( c \leq 2b \). Substituting this into the inequality, we get:
   \[
   (b + \frac{1}{6})^2 \leq \frac{2b}{3}.
   \]
   Simplifying this inequality:
   \[
   (b + \frac{1}{6})^2 \leq \frac{2b}{3} \implies b^2 + \frac{b}{3} + \frac{1}{36} \leq \frac{2b}{3}.
   \]
   Rearranging terms:
   \[
   b^2 - \frac{b}{3} + \frac{1}{36} \leq 0 \implies (b - \frac{1}{6})^2 \leq 0.
   \]
   Since the square of a real number is non-negative, the only solution is:
   \[
   b = \frac{1}{6}.
   \]

6. **Determine \( c \):**
   From \( c \leq 2b \) and \( b = \frac{1}{6} \), we get:
   \[
   c \leq 2 \cdot \frac{1}{6} = \frac{1}{3}.
   \]
   Since we have equality in the Cauchy-Schwarz inequality, \( c = \frac{1}{3} \).

7. **Form of \( g(t) \):**
   Since we have equality in the Cauchy-Schwarz inequality, \( g'(t) \) must be proportional to \( 1-t \):
   \[
   g'(t) = k(1-t).
   \]
   Integrating \( g'(t) \):
   \[
   g(t) = -\frac{1}{6} + k \left( t - \frac{t^2}{2} \right).
   \]

8. **Determine \( k \):**
   Integrate \( g(t) \) over \([0, 1]\) and set it equal to \( b \):
   \[
   \int_0^1 g(t) \, dt = \int_0^1 \left( -\frac{1}{6} + k \left( t - \frac{t^2}{2} \right) \right) dt = -\frac{1}{6} + k \left( \frac{1}{2} - \frac{1}{6} \right) = -\frac{1}{6} + \frac{k}{3}.
   \]
   Since \( b = \frac{1}{6} \):
   \[
   \frac{1}{6} = -\frac{1}{6} + \frac{k}{3} \implies \frac{k}{3} = \frac{1}{3} \implies k = 1.
   \]

9. **Final form of \( g(t) \):**
   \[
   g(t) = -\frac{1}{6} + t - \frac{t^2}{2}.
   \]

10. **Transform back to \( f(x) \):**
    Since \( g(x) = f(1-x) \):
    \[
    f(x) = g(1-x) = -\frac{1}{6} + (1-x) - \frac{(1-x)^2}{2} = 1 - x - \frac{1}{2} + x - \frac{x^2}{2} = \frac{1}{3} - \frac{x^2}{2}.
    \]

The final answer is \( \boxed{f(x) = \frac{1}{3} - \frac{x^2}{2}} \).