Limits!

$\text{We have to evaluate } \displaystyle \lim_{x \to 0^{+} } \dfrac{ e^{x^{x}-1}-x^{x} }{(x^{2x}-1)^2} \\ \displaystyle = \lim_{x \to 0^{+} } \dfrac{ e^{x^{x}-1}-x^{x} }{(x^{x}+1)^2(x^{x}-1)^2} \\ = \dfrac14\displaystyle \lim_{x \to 0^{+} } \dfrac{ e^{x^{x}-1}-x^{x} }{(x^{x}-1)^2} \\ \text{Using LH rule, we get } \\ = \dfrac14\displaystyle \lim_{x \to 0^{+} } \dfrac{ e^{x^{x}-1}\cdot x^{x}(1+\ln{x})-x^{x}(1+\ln{x}) }{2(x^{x}-1)\cdot x^{x}(1+\ln{x})} \\ = \dfrac18\displaystyle \lim_{x \to 0^{+} } \dfrac{ e^{(x^{x}-1)}-1 }{(x^{x}-1)} \\ \text{Using Formula } \displaystyle \lim_{f(x) \to 0 } \dfrac{e^{f(x)}-1}{f(x)}=1 \\ \text{Our answer is } \color{#3D99F6}{\boxed{\dfrac{1}8}}$