If $f\left(x\right)=\left\{x^2\right\}-{\left(\left\{x\right\}\right)}^2,$ where $\left\{x\right\}$ denotes the fractional part of x then

$f\left(2\right)=0$

$f\left(2^{+}\right)=\left\{4^{+}\right\}-{\left\{2^{+}\right\}}^2=0-0=0$

$f\left(2^{-}\right)=\left\{4^{-}\right\}-{\left\{2^{-}\right\}}^2=1-1=0$

Hence, $f\left(x\right)$ is continuous at x=2

$f(-2)=0$

 $f(-2^{+})=\left\{4^{-}\right\}-{\{-2^{+}\}}^2=1-0=0\ne f(-2-)$

Hence, $f\left(x\right)$ is discontinuous at x=-2