Few statements based on Continuity of functions:-
1) If $\lim_{x\to a }f(x) = b$ and $g(x)$ is continuous at $x=b$ then $\lim_{x\to a }g(f(x)) = g(b)$
2) If $\lim_{x\to a }f(x) = f(c)$ and $g(x)$ is continuous at $x=f(c) $ then $\lim_{x\to a }g(f(x)) = g(f(c))$ (in words, if f(x) is continuous at $x=c$
and g(x) is continuous at $x=f(c)$ then composition of $g(f(x))$ is also continuous at $x=c$.)
3) if $f(x)$ is continuous at $x=c$ then $\dfrac{1}{f}$ also continuous at $x=c$.(Provided that $f(c) \neq 0$).
now coming to question $f(x)= sin\left(\dfrac{1}{x^2+1}\right)$ ,
Let $g(x) = \dfrac{1}{x^2+1}$ , $h(x) = sin(x)$, now $f(x) $ is nothing but a compostion $h(g(x))$ , $x^2 +1$ is continuous everywhere because
its polynomial and observe carefully it is always positive never equals to 0, $\dfrac{1}{x^2+1}$ is also continuous everywhere (statement 3).
we all know that $sin(x)$ is continuous at every $x$, as $h(x) , g(x)$ continuous everywhere their composition $h(g(x))$ also continuous at every $x$.
so f(x) is continuous for all $x$.
$f'(x) = - 2xcos\left(\dfrac{1}{x^2+1}\right)\dfrac{1}{(x^2+1)^2}$,
now its trivial to answer the value of $f'(0)$ $= -2(0)cos(1)(1) = 0$
the answer is option C.