The representations  for the right members of (1.81) and (1.82) are obtained by the technique of pages 164-165.  Here, we merely check directly the equality of the left and right members for each of (1.81) and (1.82).

The version number of Mathematica used to evaluate this notebook:

$VersionNumber

B[1, 1][
          z_] = w[1, 1][z] - (w[0, 1][z])^2;

B[1, 2][
          z_] = w[1, 2][z] - w[0, 1][z]*w[0, 2][z];

B[2, 2][
          z_] = w[2, 2][z] - (w[0, 2][z])^2;

Inv[2, 1, 1][
          z_] = w[1, 1][z] - (w[0, 1][z])^2;

Inv[2, 1, 2][
          z_] = w[1, 2][
            z] - (1/2) w[0, 1][z]*w[1, 1][z] - w[0, 1][
              z]*w[0, 2][
              z] + (1/2) (w[0, 1][z])^3 - (1/4) 
            D[w[1, 1][z], z] + (1/2) w[0, 1][z]*
            D[ w[0, 1][z], z];

\(\(\(\(Inv[2, 2, 2]\)[
          z_] = \(w[2, 2]\)[
            z] - \(w[0, 1]\)[z]*\(w[1, 2]\)[z] + \((1/5)\) \(w[0, 2]\)[
              z]*\(w[1, 1]\)[
              z] - \((\(w[0, 2]\)[z])\)^2 + \((1/
                5)\) \((\(w[0, 1]\)[z])\)^2*\(w[1, 1]\)[
              z] + \((4/5)\) \((\(w[0, 1]\)[z])\)^2*\(w[0, 2]\)[
              z] - \((1/5)\) \((\(w[0, 1]\)[z])\)^4 - \((1/3)\) 
            D[\(w[1, 2]\)[z], z] + \((1/30)\) 
            D[\(w[1, 1]\)[z], {z, 2}] + \((1/6)\) \(w[0, 1]\)[z]*
            D[\ \(w[1, 1]\)[z], z] + \((1/15)\) 
            D[\(w[0, 1]\)[z], z]*\(w[1, 1]\)[z] + \((1/3)\) \(w[0, 1]\)[z]*
            D[\(w[0, 2]\)[z], z] + \((1/3)\) 
            D[\(w[0, 1]\)[z], z]*\(w[0, 2]\)[z] - \((1/15)\) \(w[0, 1]\)[z]*
            D[\(w[0, 1]\)[z], {z, 2}] - \((1/
                15)\) \((D[\(w[0, 1]\)[z], z])\)^2 - \((2/
                5)\) \((\(w[0, 1]\)[z])\)^2*D[\(w[0, 1]\)[z], z];\)\(
\)
    \)\)

a[2, 0, 2][
          z_] = w[0, 2][z] - (1/2) 
            D[ w[0, 1][z], 
              z] - (1/4) (w[0, 1][z])^2;

E6[z_] = B[1, 1][z]*B[2, 2][z] - 
          2  B[1, 2][z]^2 + w[0, 1][z]*B[1, 1][z]*B[1, 2][
              z] - w[0, 2][z]*B[1, 1][z]^2 + B[1, 1][z]*
            D[B[1, 2][z], z] - 
          D[B[1, 1][z], z]*B[1, 2][z];

The evaluation

Expand[
      E6[z] - ((Inv[2, 1, 1][z]*Inv[2, 2, 2][z] - 
                2  Inv[2, 1, 2][z]^2) + (4/
                  3) ( Inv[2, 1, 1][z]*
                    D[Inv[2, 1, 2][z], z] - (3/2) Inv[2, 1, 2][z]*
                    D[Inv[2, 1, 1][z], z]) + (3/
                  10)*(Inv[2, 1, 1][z]*
                    D[Inv[2, 1, 1][z], {z, 2}] - (5/4)*
                    D[Inv[2, 1, 1][z], z]^2 - 
                  4*a[2, 0, 2][
                      z]*Inv[2, 1, 1][z]^2 ))]

provides a check on the correctness of (1.81).

H7[z_] = B[1, 1][z]*D[B[2, 2][z], z] - 
          D[B[1, 1][z], z]*B[2, 2][z] - 
          2  B[1, 2][z]*B[2, 2][z] + 
          2  w[0, 1][z]*B[1, 1][z]*B[2, 2][z] - 
          2  w[0, 2][z]*B[1, 1][z]*B[1, 2][z];

E7[z_] = 
        H7[z] - (1/6) (D[E6[z], z] + 
                6  w[0, 1][z]*E6[z]);

The evaluation

Expand[
      E7[z] - (-2 Inv[2, 1, 2][z]*Inv[2, 2, 2][
                z] + (5/
                  6) ( Inv[2, 1, 1][z]*D[Inv[2, 2, 2][z], z] - 
                  2  Inv[2, 2, 2][z]*D[Inv[2, 1, 1][z], z]) + (1/
                  9) (Inv[2, 1, 1][z]*
                    D[Inv[2, 1, 2][z], {z, 2}] - (7/2) 
                    D[Inv[2, 1, 1][z], z]*
                    D[Inv[2, 1, 2][z], z] + (21/10) 
                    D[Inv[2, 1, 1][z], {z, 2}]*Inv[2, 1, 2][
                      z] - (72/5) a[2, 0, 2][z]*Inv[2, 1, 1][
                      z]*Inv[2, 1, 2][z]))]

provides a check on the correctness of (1.81).