Would substitution be helpful? That's the only thing I can think of but then I'm not sure how to derive $\ln(x^2)$
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$\begingroup$$e^{\ln(anything)} = anything$, i.e., exponention and log are inverses. So the thing you need to differentiate is just $$ x \cdot x^2 = x^3. $$ I'll bet you can differentiate that.
$\endgroup$ 1 $\begingroup$\begin{align} \frac{d}{dx}xe^{\ln x^2}&=\frac{d}{dx}x\cdot x^2 & \text{because } e^{\ln x^2}=x^2\\ &= \frac{d}{dx} x^3 & \text{exponent law: } x\cdot x^2 = x^{1+2}=x^3\\ &=3x^2 & \text{use power rule: } \frac{d}{dx} x^3=3x^{3-1}=3x^2 \end{align}
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