Differentiable bijection $f:\mathbb{R} \to \mathbb{R}$ with nonzero derivative whose inverse is not differentiable

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 $[1,2)$).

After some thoughts and discussions, I've come to the unfortunate realization that such a function ($f:\mathbb{R} \to \mathbb{R}$ a differentiable bijection, $f'>0$ WLOG (which is equivalent to $f' \neq 0$ because of Darboux's mean value theorem), $f^{-1}$ not everywhere-differentiable), does not exist, for a rather trivial reason.

If $f:\mathbb{R} \to \mathbb{R}$ were a differentiable bijection, it would be a continuous bijection, and therefore by the invariance of domain theorem, $f^{-1}$ is also continuous on all of $\mathbb{R}$, so all the conditions of the inverse funciton theorem are met, therefore $f^{-1}$ is differentiable everywhere.

PS: I'm not touting my horn by answering my own question, and zhw's answer is simpler than mine and correct, so I've accepted it. I just happened to write it out at the same time as them.