Answer
$$ y'=\frac{2\ln x}{x}e^{(\ln x)^2}.$$
Work Step by Step
Recall that $(e^x)'=e^x$
Recall that $(\ln x)'=\dfrac{1}{x}$
Recall that $(x^n)'=nx^{n-1}$
Since $ y=e^{(\ln x)^2}$, then by the chain rule, we have
$$ y'=e^{(\ln x)^2}(2\ln x)(1/x)=\frac{2\ln x}{x}e^{(\ln x)^2}.$$
