Deriving the Euler equation from a Continuous Time Dynamic Programming Problem (HJB)

As already commented, the equation you probably meant is $$ rho V(k)= sup_c {, u(c) + V'(k) ( f(k) -delta k -c ) ,}. $$ I have never seen this equation called the HJB equation (probably missing a basic reference on my part). I'll call it "dynamic programming PDE".

What you're really asking is the connection between two approaches to solve control problems---Calculus of Variations/Optimal Control vs. Dynamic Programming. Optimal Control (Pontryagin's Maximum Principle) is a first-order perturbation argument and Dynamic Programming Princple is a backward induction argument.

"Euler equation" arises from a first-order perturbation argument. (In continuous-time, it's a classical Calculus of Variation equation. In discrete time, I've only heard it used in economics, describing intertemporal consumption smoothing, but it's a perturbation argument just the same.)

In continuous-time, first-order perturbation of optimal path means a perturbation along the entire path. In contrast, in discrete time, it suffices to perturb a single period--therefore one can obtain the Euler equation by simply differentiating the Bellman equation (under e.g. Benveniste-Scheinkman assumptions that ensure differentiability of value function).

In continuous-time, I don't believe the answer is as trivial. There is, however, a classical connection between Optimal Control and Dynamic Programming via the method of characteristics. Part of the connection is that, if the value function is sufficiently smooth, then the characteristic equations of the dynamic programming PDE gives Pontryagin's Maximum Principle.

In your growth example, the maximizer of RHS is given by $u'(c^*) = V'$. Substituting then gives an implicit first order ODE $$ F(k, V, V') = rho V - u(c^*( V')) - V'cdot( f(k) -delta k) + V' cdot c^*( V') = 0. $$ Following the method of characteristics heuristicly, one of the characteristic equations would be begin{align*} frac{d}{dt} (V') &= -lambda (F_k + F_V V') &= -lambda (- V' f'(k) + V' rho), end{align*} for some $lambda geq 0$. Since $u'(c^*) = V'$, begin{align*} frac{d}{dt} log (u'(c^*)) &= -lambda (F_k + F_V V') &= lambda ( f'(k) - rho), end{align*} which is an Euler equation.

(Incidentally, the ODE $F(k, V, V')$ does not seem amenable to guess-and-verify, even when $u(c) = log c$.)