The function

is continuous at 0,

is not continuous at 0 (as you can check by computation). However,

is differentiable at 0 (

as

tends to 0)...

Your argument could be made rigorous using additional properties. You can find two directions (

for

) where different finite limits

*exist* for

. And then use the following fact: if a continuous function is differentiable on

and such that its derivative admits a finite limit

at

, then the function is differentiable at

and its derivative at

is

. But this is a bit overcomplicated.