Find the set of all quadruplets $(x,y, z,w)$ of non-zero real numbers which satisfy
$$1 +\frac{1}{x}+\frac{2(x + 1)}{xy}+\frac{3(x + 1)(y + 2)}{xyz}+\frac{4(x + 1)(y + 2)(z + 3)}{xyzw}= 0$$
1. Start with the given equation:
   \[
   1 + \frac{1}{x} + \frac{2(x + 1)}{xy} + \frac{3(x + 1)(y + 2)}{xyz} + \frac{4(x + 1)(y + 2)(z + 3)}{xyzw} = 0
   \]

2. Combine the terms to simplify the expression:
   \[
   1 + \frac{1}{x} + \frac{2(x + 1)}{xy} + \frac{3(x + 1)(y + 2)}{xyz} + \frac{4(x + 1)(y + 2)(z + 3)}{xyzw}
   \]

3. Factor out the common term \((x + 1)(y + 2)\):
   \[
   1 + \frac{1}{x} + \frac{2(x + 1)}{xy} + \frac{3(x + 1)(y + 2)}{xyz} + \frac{4(x + 1)(y + 2)(z + 3)}{xyzw} = \frac{(x + 1)(y + 2)}{xy} \left[1 + \frac{3}{z} + \frac{4(z + 3)}{zw}\right]
   \]

4. Simplify the expression inside the brackets:
   \[
   1 + \frac{3}{z} + \frac{4(z + 3)}{zw} = 1 + \frac{3}{z} + \frac{4z + 12}{zw} = 1 + \frac{3}{z} + \frac{4}{w} + \frac{12}{zw}
   \]

5. Combine the terms:
   \[
   \frac{(x + 1)(y + 2)}{xy} \left[1 + \frac{3}{z} + \frac{4}{w} + \frac{12}{zw}\right] = \frac{(x + 1)(y + 2)(z + 3)(w + 4)}{xyzw}
   \]

6. Set the equation to zero:
   \[
   \frac{(x + 1)(y + 2)(z + 3)(w + 4)}{xyzw} = 0
   \]

7. Since the denominator \(xyzw\) is non-zero, the numerator must be zero:
   \[
   (x + 1)(y + 2)(z + 3)(w + 4) = 0
   \]

8. Therefore, the solutions are:
   \[
   x = -1 \quad \text{or} \quad y = -2 \quad \text{or} \quad z = -3 \quad \text{or} \quad w = -4
   \]

The final answer is \( \boxed{ (x, y, z, w) } \) such that \(x = -1\) or \(y = -2\) or \(z = -3\) or \(w = -4\).