Given function 

$f\left(x\right)=\left[\begin{array}{cc}\frac{\sin \left(x^2\right)}{x}& , x\ne 0\\ 0& , \mathrm{if} x=0\end{array}\right.$

then $\underset{x\to 0}{\lim }f\left(x\right)=\underset{x\to 0}{\lim }\frac{\sin x^2}{x}=\underset{x\to 0}{\lim }$

$x\frac{\sin x^2}{x^2}=0=f\left(0\right)$

Hence, $f\left(x\right)$ is continuous at $x=0$

Now, for differentiability 

$\mathrm{RHD}$ (at $x=0$)

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

and $\mathrm{LHD}$ (at $x=0$)

$=\underset{h\to 0}{\lim }\frac{f\left(0-h\right)-f\left(0\right)}{-h}=\underset{h\to 0}{\lim }\frac{\sin h^2}{h^2}=1$

So, $f\left(x\right)$ is differentiable at $x=0$

$\therefore f^{\text{'}}\left(x\right)=\left[\begin{array}{cc}2\cos \left(x^2\right)-\frac{\sin x^2}{x^2}& , \mathrm{if} x\ne 0\\ 1& , \mathrm{if} x=0\end{array}\right]$

$\because \underset{x\to 0}{\lim }{f}^{\text{'}}\left(x\right)=\underset{x\to 0}{\lim }\left[2\cos \left(x^2\right)-\frac{\sin x^2}{x^2}\right]$

$=2-1=1$

$\therefore \underset{x\to 0}{\lim }{f}^{\text{'}}\left(x\right)=f^{\text{'}}\left(0\right)$

So, $f\left(x\right)$ is differentiable and the derivative is continuous.