\( \left. \begin{array} { l } { f ( x ) = \frac { x ^ { 2 } } { e ^ { x } } } \\ { f ^ { \prime } ( x ) = \square } \end{array} \right. \)

To find the derivative \( f'(x) \) of the function \( f(x) = \frac{x^2}{e^x} \), we can apply the quotient rule. The quotient rule states that if you have a function that is the ratio of two functions \( u(x) \) and \( v(x) \), such that \( f(x) = \frac{u(x)}{v(x)} \), then the derivative is given by: \[ f'(x) = \frac{u'(x)v(x) - u(x)v'(x)}{(v(x))^2} \] In this case, let \( u(x) = x^2 \) and \( v(x) = e^x \). Then, we compute their derivatives: \( u'(x) = 2x \) and \( v'(x) = e^x \). Plugging these into the quotient rule gives us: \[ f'(x) = \frac{(2x)(e^x) - (x^2)(e^x)}{(e^x)^2} \] Simplifying this, we can factor out \( e^x \) from the numerator: \[ f'(x) = \frac{e^x(2x - x^2)}{e^{2x}} = \frac{2x - x^2}{e^x} \] So, the completed derivative is: \[ f'(x) = \frac{2x - x^2}{e^x} \]