Let $ f_m: \mathbb R \to \mathbb R$, $ f_m(x)\equal{}(m^2\plus{}m\plus{}1)x^2\minus{}2(m^2\plus{}1)x\plus{}m^2\minus{}m\plus{}1,$ where $ m \in \mathbb R$.

1) Find the fixed common point of all this parabolas.

2) Find $ m$ such that the distance from that fixed point to $ Oy$ is minimal.
1. **Finding the fixed common point of all these parabolas:**

   We start with the given function:
   \[
   f_m(x) = (m^2 + m + 1)x^2 - 2(m^2 + 1)x + (m^2 - m + 1)
   \]

   To find the fixed point, we need to find a value of \( x \) such that \( f_m(x) = x \) for all \( m \). This means solving the equation:
   \[
   (m^2 + m + 1)x^2 - 2(m^2 + 1)x + (m^2 - m + 1) = x
   \]

   Rearrange the equation:
   \[
   (m^2 + m + 1)x^2 - 2(m^2 + 1)x + (m^2 - m + 1) - x = 0
   \]

   Combine like terms:
   \[
   (m^2 + m + 1)x^2 - (2m^2 + 2 + 1)x + (m^2 - m + 1) = 0
   \]
   \[
   (m^2 + m + 1)x^2 - (2m^2 + 3)x + (m^2 - m + 1) = 0
   \]

   This is a quadratic equation in \( x \). To find the fixed point, we need to solve for \( x \) using the quadratic formula:
   \[
   x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \( a = m^2 + m + 1 \), \( b = -(2m^2 + 3) \), and \( c = m^2 - m + 1 \).

   Substitute these values into the quadratic formula:
   \[
   x = \frac{2m^2 + 3 \pm \sqrt{(2m^2 + 3)^2 - 4(m^2 + m + 1)(m^2 - m + 1)}}{2(m^2 + m + 1)}
   \]

   Simplify the discriminant:
   \[
   (2m^2 + 3)^2 - 4(m^2 + m + 1)(m^2 - m + 1)
   \]
   \[
   = 4m^4 + 12m^2 + 9 - 4(m^4 + m^2 - m + m^2 - m + 1)
   \]
   \[
   = 4m^4 + 12m^2 + 9 - 4m^4 - 4m^2 + 4m - 4m^2 + 4m - 4
   \]
   \[
   = 4m^2 + 8m + 5
   \]

   Substitute back into the quadratic formula:
   \[
   x = \frac{2m^2 + 3 \pm \sqrt{4m^2 + 8m + 5}}{2(m^2 + m + 1)}
   \]

   Notice that the discriminant simplifies to:
   \[
   \sqrt{4(m^2 + 2m + 1)} = \sqrt{4(m + 1)^2} = 2(m + 1)
   \]

   Therefore:
   \[
   x = \frac{2m^2 + 3 \pm 2(m + 1)}{2(m^2 + m + 1)}
   \]

   This gives us two solutions:
   \[
   x = \frac{2m^2 + 3 + 2m + 2}{2(m^2 + m + 1)} = \frac{2m^2 + 2m + 5}{2(m^2 + m + 1)}
   \]
   \[
   x = \frac{2m^2 + 3 - 2m - 2}{2(m^2 + m + 1)} = \frac{2m^2 - 2m + 1}{2(m^2 + m + 1)}
   \]

   Simplify each:
   \[
   x = \frac{2(m^2 + m + 1) + 3}{2(m^2 + m + 1)} = 1 + \frac{3}{2(m^2 + m + 1)}
   \]
   \[
   x = \frac{2(m^2 - m + 1)}{2(m^2 + m + 1)} = \frac{m^2 - m + 1}{m^2 + m + 1}
   \]

   Since we are looking for a fixed point common to all parabolas, we need \( x \) to be independent of \( m \). The only solution that fits this criterion is:
   \[
   x = 1
   \]

   Therefore, the fixed point is \( (1, 0) \).

2. **Finding \( m \) such that the distance from the fixed point to the y-axis is minimal:**

   The distance from the fixed point \( (1, 0) \) to the y-axis is always 1, regardless of \( m \). Therefore, there is no specific \( m \) that minimizes this distance further.

The final answer is \( \boxed{ (1, 0) } \)