Answer
$$ f'(x)= \frac{(2x^2 -1)e^{x^2}}{x^2}.$$
Work Step by Step
Recall the quotient rule: $(\dfrac{u}{v})'=\dfrac{vu'-uv'}{v^2}$
Recall that $(e^x)'=e^x$
Since we have
$$ f(x)=\frac{e^{x^2}}{x}$$
then the derivative $ f'(x)$, using the chain and quotient rules, is given by
$$ f'(x)=\frac{x e^{x^2}(x^2)'-e^{x^2}}{x^2}=\frac{2x^2 e^{x^2}-e^{x^2}}{x^2}=\frac{(2x^2 -1)e^{x^2}}{x^2}.$$
