The map

f[{x_, y_}]  =  {x^3 - 2  x^2 y + y, y^3 - 2  x^2}

The derivative

Df[{x_, 
            y_}]  =  {{3  x^2 - 4  x y,  -2 x^2  +  1}, {-4 x, 
            3  y^2}};
MatrixForm[Df[{x, y}]]

The linear approximation about p evaluated at q

L[p_, q_]  :=  Simplify[f[p] + Df[p] . (q - p)]

The initial point, it's image, and a check to see that Df is invertible at this point

guess  =  {1, 1}
f[guess]
Det[Df[guess]]

Finding a better guess

L[guess, {x, y}]
guess  =  {x, y} /. 
        First[Solve[
            L[guess, {x, y}] == { .1, - .8}, {x, 
              y}]]
f[guess]

Finding a better guess

L[guess, {x, y}]
guess = {x, y} /. 
        First[Solve[
            L[guess, {x, y}] == { .1, - .8}, {x, y}]]
f[guess]

Finding a better guess

L[guess, {x, y}]
guess = {x, y} /. 
        First[Solve[
            L[guess, {x, y}] == { .1, - .8}, {x, y}]]
f[guess]