Answer
$$ y' =x(\ln x^2 + 1)$$
Work Step by Step
Recall the product rule: $(uv)'=u'v+uv'$
Recall that $(\ln x)'=\dfrac{1}{x}$
Recall that $(x^n)'=nx^{n-1}$
Since $ y=x^2 \ln x $, then using the product rule we have
$$ y'=2x \ln x+ x^2 \frac{1}{x}=2x \ln x +x=x(\ln x^2 + 1)$$
