Answer
$$ f'(x)= -\frac{1}{x^2}e^{\frac{1}{x}}.$$
Work Step by Step
Recall that $(e^x)'=e^x$
Recall that $(x^n)'=nx^{n-1}$
Since we have
$$ f(x)= e^{\frac{1}{x}}$$
then the derivative $ f'(x)$, using the chain rule, is given by
$$ f'(x)= e^{\frac{1}{x}}\left(\frac{1}{x}\right)'=-\frac{1}{x^2}e^{\frac{1}{x}}.$$
