$\underset{x\to 0^{+}}{\lim }{x}^{x^2}=e^{\underset{x\to 0^{+}}{\lim }{x}^2\ln x}=e^L$ (say)
$L=\underset{x\to 0^{+}}{\lim }\frac{\ln x}{\frac{1}{x^2}}=\underset{x\to 0^{+}}{\lim }\frac{{\displaystyle \frac{1}{x}}}{\left(-\frac{2}{x^3}\right)}$ {applying L’ hospital rule}
$=\underset{x\to 0^{+}}{\lim }\left(-\frac{x^2}{2}\right)=0$
Hence, $\underset{x\to 0^{+}}{\lim }{x}^{x^2}=e^0=1$

Similarly, $\underset{x\to 0^{+}}{\lim }{x}^x=e^{\underset{x\to 0^{+}}{\lim }x\ln x}=e^{\underset{x\to 0^{+}}{\lim }\frac{\ln x}{{\displaystyle \frac{1}{x}}}}$

$=e^{\underset{x\to 0^{+}}{\lim }\frac{{\displaystyle \frac{1}{x}}}{{\displaystyle \frac{-1}{x^2}}}}=e^{\underset{x\to 0^{+}}{\lim }-x}=e^0=1$

Hence, $\underset{x\to 0^{+}}{\lim }{x}^{\left(x^x\right)}=0$

Therefore, $\underset{x\to 0^{+}}{\lim }{x}^{x^2}+\underset{x\to 0^{+}}{\lim }{x}^{\left(x^x\right)}=1$