To check differentiable at x = 0

$$R\{ f'(0)\} = \mathop {\lim }\limits_{h \to 0} {{f(0 + h) - f(0)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{{h^2}\left| {\cos {\pi \over h}} \right| - 0} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} h.\,\left| {\cos {\pi \over h}} \right| = 0$$

$$L\{ f'(0)\} = \mathop {\lim }\limits_{h \to 0} {{f(0 - h) - f(0)} \over { - h}}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{{h^2}\left| {\cos \left( { - {\pi \over h}} \right)} \right| - 0} \over { - h}}$$

$$ = 0$$

So, f(x) is differentiable at x = 0

To check differentiability at x = 2

$$R\{ f'(2)\} = \mathop {\lim }\limits_{h \to 0} {{f(2 + h) - f(2)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}\left| {\cos \left( {{\pi \over {2 + h}}} \right)} \right| - 0} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\cos \left( {{\pi \over {2 + h}}} \right)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\sin \left( {{\pi \over 2} - {\pi \over {2 + h}}} \right)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 + h)}^2}.\sin \left( {{{\pi h} \over {2(2 + h)}}} \right)} \over {h.{\pi \over {2(2 + h)}}.{\pi \over {2(2 + h)}}}}$$

$$ \Rightarrow R\left( {f'(2)} \right) = \pi $$

$$L\left( {f'(2)} \right) = \mathop {\lim }\limits_{h \to 0} {{f(2 - h) - f(2)} \over { - h}}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2}.\left| {\cos {\pi \over {2 - h}}} \right| - {2^2}.\left| {\cos {\pi \over 2}} \right|} \over { - h}}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2} - \left( { - \cos {\pi \over {2 - h}}} \right) - 0} \over { - h}}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{ - {{(2 - h)}^2}.\sin \left( {{\pi \over 2} - {\pi \over {2 - h}}} \right)} \over h}$$

$$ = \mathop {\lim }\limits_{h \to 0} {{{{(2 - h)}^2}.\sin \left( { - {{\pi h} \over {2(2 - h)}}} \right)} \over {h \times {{ - \pi } \over {2(2 - h)}} \times {{ - \pi } \over {2(2 - h)}}}}$$

$$ \Rightarrow L(f'(2)) = - \pi $$

Thus, f(x) is differentiable at x = 0 but not at x = 2.