Is it valid to derive Fenchel conjugate for a non-smooth function? Checking its definition $f^*(y) = sup_{x \in \mathsf{dom}f} (y^Tx - f(x))$, I think this would be OK, but I'm not sure about that.
An example is the (one-dimensional) bottom-up-triangle-shaped function with its domain the interval $[0,1]$. Can we simply say its conjugate is the hinge-shaped function (which is by definition if its conjugate is allowed)?
Yes, the definition is valid for any convex function. When the domain of $f$ is unbounded, the supremum may turn out to be infinite, but even this is not a big deal: $+\infty$ is routinely allowed as a value in convex analysis. In fact, the (Legendre-)Fenchel transform makes sense for any real-valued function whatsoever, it does not even have to be convex. But the word conjugate is used only when $f$ is convex, because then (under some technical assumption) we have $f^{**}=f$.
For example, if $f(x)=\max(2x,3x)$ on $[-1,1]$, then the Fenchel transform is $$f^*(x)=\begin{cases} -y-2,\quad &y\le -2 \\ 0 ,\quad &-2\le y\le 3 \\ y-3,\quad & y \ge 3 \end{cases} $$