Split into right and left limits.

Right limit ($x\to0^+$): $[x]=0$, so $\sin(x[x])=0$, $|x|=x$.

$\dfrac{\tan(\pi\sin^2x)+x^2}{x^2}$. Using $\tan(\pi\sin^2x)\approx\pi\sin^2x\approx\pi x^2$:

$=\pi+1$

Left limit ($x\to0^-$): $[x]=-1$, $\sin(x[x])=\sin(-x)=-\sin x$, $|x|=-x$.

$=\dfrac{\tan(\pi\sin^2x)+(-x-(-\sin x))^2}{x^2}=\dfrac{\pi x^2+(\sin x-x)^2}{x^2}\to\pi+0=\pi$

Left $\neq$ Right, so the limit does not exist.