Let $$f(x) = \begin{cases} (x-2)\cos(\frac{1}{x-2}), & \text{if $x\neq$ 0} \\[2ex] 0, & \text{if $x$=0} \end{cases}$$ And we have to check for the differentiability at the points $x=0$ and $x=2$.
We can easily check the differentiability by using the formula $f '(a)=\lim\limits_{h \to a} \frac{f(a+h)-f(a)}{h} $ and surely the answer checks out the $f(x)$ is differentiable at $x=0$ but not at $x=2$.
But instead of solving for the whole function, I went up to solve it in a different way, I checked the differentiability of the $f(x)=x-2$ at the points $x=0$ and $x=2$ and turned out that the function is differentiable at both the points. So, I tried a new trick here, that if there are two different function in multiplication ( $h(x)= f(x)\cdot g(x)$ ) such that $f(x)$ is differentiable at a point $x=a$, then surely the whole function( $h(x)= f(x)\cdot g(x)$ ) is differentiable at the point $x=a$.
This seems to work for other functions such as checking the differentiability of $f(x)=x|x|$ at $x=0$.
I am not sure why it is not working in the first one.