Answer
$$y'=\frac{-e^{\frac{1}{x}}-2xe^{\frac{1}{x}} }{x^4}$$
Work Step by Step
Given $$y=\frac{e^{\frac{1}{x}}}{x^2}$$
Then
\begin{align*}
y'&=\frac{\frac{d}{dx}\left(e^{\frac{1}{x}}\right)x^2-\frac{d}{dx}\left(x^2\right)e^{\frac{1}{x}}}{\left(x^2\right)^2}\\
&=\frac{\left(-\frac{e^{\frac{1}{x}}}{x^2}\right)x^2-2xe^{\frac{1}{x}}}{\left(x^2\right)^2}\\
&=\frac{-e^{\frac{1}{x}}-2xe^{\frac{1}{x}} }{x^4}
\end{align*}
