If $f\left(x\right)=\left\{\begin{array}{cl}\frac{x^2\ln \cos x}{\ln \left(1+x^2\right)}& , x\ne 0\\ 0& , x=0\end{array}\right.$, then $f\left(x\right)$ is

$f^{\text{'}}\left(0^{+}\right)=\underset{h\to 0}{\lim }\frac{f\left(0+h\right)-f\left(0\right)}{h}$

$=\underset{h\to 0}{\lim }\frac{f\left(h\right)}{h}$

$=\underset{h\to 0}{\lim }\frac{\ln \cos h}{\ln \left(1+h^2\right)}\times \frac{h^2}{h^2} ....\left(1\right)$   {Multiplying Numerator & denominator by $h$}

$=\underset{h\to 0}{\lim }\frac{h^2}{\ln \left(1+h^2\right)}\times \underset{h\to 0}{\lim }\frac{\ln \cos \mathit{h}}{h^2}$

$=\underset{h\to 0}{\lim }\frac{h^2}{\ln \left(1+h^2\right)}\times \underset{h\to 0}{\lim }-\frac{\tan \mathit{h}}{2h}$      {Using L'Hospital's Rule}
 $=1\times -\frac{1}{2}$
$=-\frac{1}{2}$

and   $f^{\text{'}}\left(0^{-}\right)=\underset{h\to 0}{\lim }\frac{f\left(0-h\right)-f\left(0\right)}{-h}$

$=\underset{h\to 0}{\lim }\frac{f\left(-h\right)}{-h}$

$=-\underset{h\to 0}{\lim }\frac{-h\ln cos\left(-h\right)}{\ln \left(1+{\left(-h\right)}^2\right)\left(-h\right)}$

$=\underset{h\to 0}{\lim }\frac{\mathrm{logcos}\mathit{h}}{\log \left(1+h^2\right)}$

$=-\frac{1}{2}$                                 {from Eq. $\left(1\right)$}

$\therefore f^{\text{'}}\left(0^{+}\right)=f^{\text{'}}\left(0^{-}\right)$

$\therefore$ $f\left(x\right)$ is differentiable and hence continuous at $x=0$.