Find the greatest and smallest value of $f(x, y) = y-2x$, if x, y are distinct non-negative real numbers with $\frac{x^2+y^2}{x+y}\leq 4$.
1. **Rewrite the given inequality:**
   \[
   \frac{x^2 + y^2}{x + y} \leq 4
   \]
   Multiply both sides by \(x + y\) (assuming \(x + y \neq 0\) since \(x\) and \(y\) are non-negative and distinct):
   \[
   x^2 + y^2 \leq 4(x + y)
   \]

2. **Substitute \(y = 2x + z\) into the inequality:**
   \[
   x^2 + (2x + z)^2 \leq 4(x + 2x + z)
   \]
   Simplify the left-hand side:
   \[
   x^2 + 4x^2 + 4xz + z^2 \leq 4(3x + z)
   \]
   Combine like terms:
   \[
   5x^2 + 4xz + z^2 \leq 12x + 4z
   \]

3. **Rearrange the inequality:**
   \[
   5x^2 + 4xz + z^2 - 12x - 4z \leq 0
   \]

4. **Analyze the quadratic inequality:**
   This is a quadratic in \(x\):
   \[
   5x^2 + (4z - 12)x + (z^2 - 4z) \leq 0
   \]
   For this quadratic to have real solutions, its discriminant must be non-negative:
   \[
   (4z - 12)^2 - 4 \cdot 5 \cdot (z^2 - 4z) \geq 0
   \]
   Simplify the discriminant:
   \[
   (4z - 12)^2 - 20(z^2 - 4z) \geq 0
   \]
   \[
   16z^2 - 96z + 144 - 20z^2 + 80z \geq 0
   \]
   \[
   -4z^2 - 16z + 144 \geq 0
   \]
   \[
   z^2 + 4z - 36 \leq 0
   \]

5. **Solve the quadratic inequality:**
   \[
   z^2 + 4z - 36 = 0
   \]
   Use the quadratic formula \(z = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\):
   \[
   z = \frac{-4 \pm \sqrt{16 + 144}}{2}
   \]
   \[
   z = \frac{-4 \pm \sqrt{160}}{2}
   \]
   \[
   z = \frac{-4 \pm 4\sqrt{10}}{2}
   \]
   \[
   z = -2 \pm 2\sqrt{10}
   \]
   Therefore, the roots are:
   \[
   z_1 = -2 + 2\sqrt{10}, \quad z_2 = -2 - 2\sqrt{10}
   \]

6. **Determine the range of \(z\):**
   The quadratic inequality \(z^2 + 4z - 36 \leq 0\) holds between the roots:
   \[
   -2 - 2\sqrt{10} \leq z \leq -2 + 2\sqrt{10}
   \]

Conclusion:
The greatest value of \(f(x, y) = y - 2x\) is \(-2 + 2\sqrt{10}\) and the smallest value is \(-2 - 2\sqrt{10}\).

The final answer is \(\boxed{[-2 - 2\sqrt{10}, -2 + 2\sqrt{10}]}\)