\( \frac { \sqrt[ 3 ] { x ^ { 2 } } + 2 \sqrt[ 3 ] { x } + 1 } { x + 3 \sqrt[ 3 ] { x ^ { 2 } } + 3 \sqrt[ 3 ] { x } + 1 } - \frac { 1 } { \sqrt[ 3 ] { x } + 1 } \)

To simplify the expression \[ \frac { \sqrt[3]{x^2} + 2 \sqrt[3]{x} + 1 } { x + 3 \sqrt[3]{x^2} + 3 \sqrt[3]{x} + 1 } - \frac { 1 } { \sqrt[3]{x} + 1 }, \] we start by recognizing that the numerator and denominator of the first fraction resemble binomial cubes. The term \( \sqrt[3]{x^2} + 2\sqrt[3]{x} + 1 \) can be expressed as \( (\sqrt[3]{x} + 1)^2 \) because of the identity \( a^2 + 2ab + b^2 = (a + b)^2 \). Similarly, the denominator \( x + 3 \sqrt[3]{x^2} + 3 \sqrt[3]{x} + 1 \) can be expressed as \( (\sqrt[3]{x} + 1)^3 \). Thus, we rewrite the expression as: \[ \frac {(\sqrt[3]{x} + 1)^2}{(\sqrt[3]{x} + 1)^3} - \frac {1}{\sqrt[3]{x} + 1}. \] Now, simplifying the fraction gives: \[ \frac {1}{\sqrt[3]{x} + 1} - \frac {1}{\sqrt[3]{x} + 1} = 0. \] Therefore, the overall simplified result of the given expression is \[ \boxed{0}. \]