This question is a more general problem but for simplicity let us consider the particular planar dynamical system, which essentially represents a continuous gradient descent of $f$ under constraint $y = 0$, $$ \left\{ \begin{aligned} \dot x &= -\frac{\partial f}{\partial x}(x,y) \\ \dot y &= -y, \end{aligned} \right. $$ where $f$ is twice continuously differentiable and whenever $\frac{\partial f}{\partial x} (x,0) = 0$, $\frac{\partial^2 f}{\partial x^2} (x,0) \neq 0$. The equilibria are points $(x,0)$ such that $f(x,0) = 0$, they are hyperbolic, stable if $\frac{\partial^2 f}{\partial x^2} (x,0) > 0$, unstable otherwise. Suppose that reduced to the invariant manifold $M$ of points $(x,y)$ satisfying $y = 0$, the system (which is then $\dot x = -\frac{\partial f}{\partial x}(x,0)$) has only bounded trajectories. Then the positive limit set of any initial condition in $M$ must be non-empty. LaSalle's invariance principle states that, since $f$ decreases along trajectories of $M$, the limit set of an initial condition of $M$ can only contain points of $M$ for which $f$ is constant, in this case this means those for which $\frac{\partial f}{\partial x} (x,0) = 0$, namely the isolated equilibria. Since the limit set is a singleton as soon as it contains an isolated point, this means that all initial conditions of $M$ converge to an equilibrium, the stable manifolds of the system $\dot x = -\frac{\partial f}{\partial x}(x,0)$ form a partition of $M$.
If there is only one equilibrium, that is further stable, then clearly the stable manifold is an open containing $M$. In the general case where more equilibria may occur, is the union of the stable manifolds of all equilibria an open?