On whether "there exists a differentiable bijection $f: \mathbb{R} \to \mathbb{R}$ such that $f'(x) \neq 0$ for all $x \in \mathbb{R}$, but whose inverse $f^{-1}$ is non-differentiable at some point $y_{0} \in \mathbb{R}$": I don't think this is possible.
Proof: Recall that every continuous bijection on $\mathbb R$ has a continuous inverse. So certainly this holds for $f.$ Consider the difference quotient
$$\tag 1 \frac{f^{-1}(y) - f^{-1}(y_0)}{y-y_0} = \frac{f^{-1}(y) - f^{-1}(y_0)}{f(f^{-1}(y))- f(f^{-1}(y_0))}.$$
As $y\to y_0,$ $f^{-1}(y) \to f^{-1}(y_0).$ Thus $(1)$ converges to the familiar $\dfrac{1}{f'(f^{-1}(y_0))}$ and we're done.
Added later: With reference to the comments below, I found the following example: On $(-1,2)$ define $f(x) = x$ on $(-1,0].$ On $(0,1)$ we do something more complicated while keeping $f(x)$ trapped between $g(x) = x$ and $h(x) =x+x^2.$ In so doing we are guaranteed $f'(0)=1.$
On the interval $I_n =(1/(n+1),1/n)$ define $f$ to equal the line through $(1/(n+1), h(1/(n+1))$ and $(1/n,g(1/n)).$ Then $f(I_n) = (h(1/(n+1),g(1/n)).$ Verify that $f$ is between $g$ and $h$ on each $I_n.$ Also notice that the intervals $f(I_n)$ have gaps betweem them. For example $f(I_1) = (3/4,1),$ $f(I_2) = (4/9,1/2).$
So we've defined $f$ on $(-1,1).$ Now there is a bijection from $[1,2)$ onto all the above-mentioned gaps, i. e., onto $(0,1)\setminus \cup_{n=1}^{\infty}f(I_n).$ Define $f$ to be this bijection on $[1,2).$
Then $f$ maps $(-1,2)$ bijectively onto $(-1,1),$ $f(0)=0,$ $f'(0) = 1,$ but $f^{-1}$ fails to be continuous at $0$ (because there are sequences $\to 0^+$ that $f^{-1}$ sends to $[1,2)$).
The condition is exactly that $f'(x)$ is nonzero (or equivalently, positive) for all $x$. Indeed, if $f^{-1}$ is differentiable everywhere, then differentiating the identity $f^{-1}(f(x))=x$ gives $(f^{-1})'(f(x))f'(x)=1$ so $f'(x)$ can never be zero.
Conversely, if $y=f(x)$ and $f'(x)\neq 0$, then $f^{-1}$ is differentiable at $y$ with $(f^{-1})'(y)=1/f'(x)$ (indeed, you can prove this directly from the definition, and the difference quotients to compute $(f^{-1})'(y)$ are the reciprocals of the difference quotients for $f'(x)$). So if $f'$ is always nonzero, then $f^{-1}$ is differentiable everywhere with $(f^{-1})'(y)=1/f'(f^{-1}(y))$. If $f$ is twice continuously differentiable, we can then differentiate $(f^{-1})'$ by the chain rule to find that $f^{-1}$ is twice continuously differentiable.
Note, however, that merely assuming $f$ is differentiable and strictly increasing does not imply $f'$ is positive everywhere. For instance, consider $f(x)=x^3$, which is strictly increasing and infinitely differentiable but $f'(0)=0$ and $f^{-1}(x)=\sqrt[3]{x}$ is not differentiable at $0$.
Best Answer
The derivative of an differentiable function with a (differentiable) inverse function is not necessarily continuous.
The function $f: \Bbb R \to \Bbb R$ defined as $$ f(x) = \begin{cases} 4x + x^2 \sin(1/x) & \text{ if } x \ne 0 \, , \\ 0 & \text{ if } x = 0 \,, \end{cases} $$ is differentiable everywhere, with $$ f'(x) = \begin{cases} 4 + 2x \sin(1/x) -\cos(1/x) & \text{ if } x \ne 0 \, , \\ 4 & \text{ if } x = 0 \, , \end{cases} $$ but $f'$ is not continuous at $x=0$.
For all $x \ne 0$ is $|x \sin(1/x)| \le 1$ and $|\cos(1/x)| \le 1$, so that $f'(x) \ge 1$ for all $x \in \Bbb R$. It follows that $f$ is strictly increasing and therefore invertible. The inverse function is differentiable everywhere by the inverse function rule.