Question
Is there a differentiable and injective function with the discontinuous derivative?
If $f$ is differentiable then it is continuous and being continuous and one-to-one is strictly monotonous, by the intermediate value theorem (how would you prove it, by the way?) and then the function $f(x)=x²\sin(1/x)$ does not work
Define $\psi(x)= \exp\Bigl(\frac{-1}{x(1-x)}\Bigr)$ for $x \in (0,1)$ and $\psi(x)=0$ for $x \notin (0,1)$. Then $\psi$ is a $C^\infty$ bump function which vanishes with all its derivatives at $0$ and at $1,$ and satisfies $\psi(1/2)=e^{-4}$.
Next, define $\varphi:[-1,1] \to [0,\infty)$ as follows: $\:$ First, $\varphi(0):=0$. Second, $$\forall k \ge 1, \quad \forall x\in (2^{-k},2^{1-k}], \quad \text{let} \quad \varphi(x)=1+ \cdot\psi(2^k x-1) \,, $$ so that $$\forall k \ge 1, \quad \varphi(2^{-k})=1 \quad \text{and} \quad \varphi(3\cdot2^{-k-1})=1+ e^{-4} \,, \tag{*}$$ This defines $\varphi$ on $[0,1]$. Finally, for $x \in [-1,0]$ define $\varphi(-x)=-\varphi(x)$.
Now we let $\Phi(x)=\int_{-1}^x \varphi(t) \, dt$. Then $\Phi$ is a differentiable and injective function on $(-1,1)$, yet $\Phi'=\varphi$ is discontinuous at $0$.
