At $x=0$, the function takes $0/0$ form. So, the function is not continuous at $x=0$.
By mentioning $f(x)=1$ at $x=0$, an attempt has been made to make the function continuous. So, let us check whether at $\text{lim }x\rightarrow 0$, the function value equals $1$ ?
Using L'Hospital's rule, it becomes $\frac{1+e^x}{x+e^x}$ and putting $x=0$, it is $2$.
So, even after adding $f(0)=1$, the function couldn't be made continuous.
Hence, the function is not differentiable too.
Answer is d.
