Answer
$$0.04$$
Work Step by Step
Since $$ f(x) =x^{1 / 3} e^{x-1} ,\ \ \ \ \Delta x= 0.03,\ \ a=1 $$
and
\begin{align*}
f'(x)& = \frac{d}{dx}\left(x^{\frac{1}{3}}\right)e^{x-1}+\frac{d}{dx}\left(e^{x-1}\right)x^{\frac{1}{3}}\\
&= \frac{e^{x-1}+3xe^{x-1}}{3x^{\frac{2}{3}}}\\
f'(1)&=\frac{4}{3}
\end{align*}
Then
\begin{align*}
\Delta f&=f(a+\Delta x)-f(a)\\
&\approx f'(a) \Delta x\\
&\approx \frac{4(0.03)}{3}\\
&\approx 0.04
\end{align*}
