Derivative of ${x^{x^2}}$
Yes, you've done it correctly so far, and could proceed by the chain rule.
On the other hand, set
$$y = x^{x^2}$$
Then $\ln{y} = x^2 \ln{x}$; taking a derivative on both sides and using the chain rule for the left leads to
$$\frac{y'}{y} = 2x \ln{x} + \frac{x^2}{x}$$
Hint:
$$\frac{d}{dx} (e^{{x^2}log(x)})=\frac{d}{dx}\left({{x^2}log(x)}\right)e^{{x^2}log(x)}$$
$$\frac{d}{dx} (e^{{x^2}log(x)})=\frac{d}{dx}\left({{x^2}log(x)}\right)e^{{x^2}log(x)}=\left(x+2x\log x \right)(e^{{x^2}log(x)})$$