Is $\int_0^x\left|\sin\left(\frac{1}{t}\right)\right|\mathrm{d}t$ differentiable at $0$? Let $f(x)=\int_0^x\left|\sin\left(\frac{1}{t}\right)\right|\mathrm{d}t$ for $x\in\mathbb{R}$. Is $f$ differentiable at $x=0$?
This is not a routine problem. And the solution is tricky (I have seen it on MSE but my search capabilities are limited so I repeat the same here). Let me know if you need more details for any of the steps involved.

Consider the limit $$L=\lim_{x\to 0}\frac{1}{x}\int_{0}^x|\sin(1/t)|\,dt\tag{1}$$ By definition the given function is differentiable at $0$ if and only if the above limit exists. Putting $x=1/u$ and considering first $x\to 0^{+}$ we see that the above limit $L$ equals $$\lim_{u\to\infty} u\int_{0}^{1/u}|\sin(1/t)|\,dt=\lim_{u\to\infty} u\int_{u} ^{\infty} \frac{|\sin t|} {t^2}\,dt$$ Now split the interval of integration into an infinite number of intervals of type $[u+k\pi, u+(k+1)\pi]$ and thus we have $$L=\lim_{u\to\infty} u\sum_{k=0}^{\infty} \int_{u+k\pi} ^{u+(k+1)\pi}\frac{|\sin t|} {t^2}\,dt=\lim_{u\to\infty} u\sum_{k=0}^{\infty} I_k\tag{2}$$ Next use the inequality $$(u+k\pi) ^2\leq t^2\leq (u+(k+1)\pi)^2$$ to estimative the integral $I_k$ above as $$\frac{1}{(u+(k+1)\pi)^2}\int_{u+k\pi}^{u+(k+1)\pi}|\sin t|\, dt\leq\int_{u+k\pi} ^{u+(k+1)\pi}\frac{|\sin t|} {t^2}\,dt\leq\frac{1}{(u+k\pi)^2}\int_{u+k\pi}^{u+(k+1)\pi}|\sin t|\, dt$$ Thus we have $$\frac{2}{(u+(k+1)\pi)^2}\leq I_k\leq\frac{2}{(u+k\pi)^2}\tag{3}$$ We next need to estimate these lower and upper bounds using $$\frac{1}{(u+(k+1)\pi)^2}\geq \frac{1}{\pi}\int_{u+(k+1)\pi}^{u+(k+2)\pi}\frac{dt}{t^2}$$ Adding these from $k=0$ to $k=\infty$ we get $$\sum_{k=0}^{\infty} \frac{2}{(u+(k+1)\pi)^2}\geq \frac{2}{\pi}\int_{u+\pi}^{\infty} \frac{dt} {t^2}=\frac{2}{\pi(u+\pi)}\tag{4}$$ Similarly the inequality $$\frac{1}{(u+k\pi)^2}\leq \frac{1}{\pi}\int_{u+(k-1)\pi}^{u+k\pi}\frac{dt}{t^2}$$ gives us $$\sum_{k=0}^{\infty} \frac{2}{(u+k\pi)^2}\leq\frac{2}{\pi}\int_{u-\pi}^{\infty} \frac{dt} {t^2}=\frac{2}{\pi(u-\pi)}\tag{5}$$ From $(3),(4),(5)$ it follows that $$\frac{2}{\pi(u+\pi)}\leq\sum_{k=0}^{\infty} I_k\leq\frac {2}{\pi(u-\pi)}$$ Multiplying by $u$ and using Squeeze theorem we see that the limit $L$ in $(2)$ equals $2/\pi$.

The case for $x\to 0^{-}$ can be dealt by a simple substitution $x=-y$. Thus the derivative of given function at $0$ is $L=2/\pi$.
At the given formula $ f(x)=\int _{ 0 }^{ x }{ \left| \sin { \frac { 1 }{ t }  }  \right| dt }  $,
we can evaluate whether f '(0) exists.

Then we can think this problem with the definition of derivative.
$$\lim _{ h\rightarrow 0 }{ \frac { f(x+h)-f(h) }{ h }  } =\lim _{ h\rightarrow 0 }{ \frac { \int _{ 0 }^{ x+h }{ \left| \sin { \frac { 1 }{ t }  }  \right| dt\quad -\int _{ 0 }^{ x }{ \left| \sin { \frac { 1 }{ t }  }  \right| dt\quad  } \quad  }  }{ h }  } =\lim _{ h\rightarrow 0 }{ \frac { \int _{ x }^{ x+h }{ \left| \sin { \frac { 1 }{ t }  }  \right| dt\quad  }  }{ h }  } $$

Since we evaluate the function when x=0, so it is same as evaluating $$ \lim _{ h\rightarrow 0 }{ \frac { \int _{ 0 }^{ h }{ \left| \sin { \frac { 1 }{ t }  }  \right| dt\quad  }  }{ h }  } $$

The value of $ \left| \sin { \frac { 1 }{ x }  }  \right| $ does not exist when $x\rightarrow 0$.

Substitue 1/x=u, then $ lim_{ x\rightarrow 0 }{ \left| \sin { \frac { 1 }{ x }  }  \right| = }lim_{ u\rightarrow \infty  }{ \left| \sin { u }  \right|  } $, it oscillates between 0 and 1.

So the limit does not exist, therefore it is not differentiable.