if  $f\left(x\right)=\{\begin{array}{l}\frac{x{e}^{\frac{1}{x}}}{1+e^{\frac{1}{x}}}, if x\ne 0\\ 0, if x=0\end{array}$ then 

$f\left(x\right)$ is discontinuous at  $x=0$

$f\left(x\right)$ has right derivative ‘0’ at  $x=0$

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

$f\left(x\right)$ has left derivative ‘1’ at  $x=0$