Definition: An oscillatory solution is one where $(x(t), y(t))$ is a trajectory and $x(t)$ and $y(t)$ are not constant. Further, for any $n \in \mathbb{N}$ we have $x(t+nt) = x(t)$ and $y(t+nt) = y(t)$. $T$ is the period of oscillation.
Suppose we have a system
$$\dot{x} = f(x)$$ $$\dot{y} = g(y)$$
Further suppose $f(x)$ and $g(y)$ are both continuous and depend solely on their respective independent variables.
Can this system produce an oscillatory solution as defined above?
My initial thought is yes. If those derivatives can be configured such that they switch from positive to negative at favorable points, that should produce an oscillation as required above. But I can't seem to come up with a concrete $f(x)$ and $g(y)$ to prove it. Any ideas?
In your system, the dynamics of $x$ and $y$ are not coupled. For a solution $(x(t),y(t))$ to be periodic in $t$, obviously both $x$ and $y$ need to be periodic. Therefore, it suffices to answer the question for the one-dimensional ODE $\dot{x} = f(x)$.
If $x(t)$ solves the ODE $\dot{x} = f(x)$ with $f$ continuous, $x(t)$ is at least continuously differentiable in $t$. Suppose $x_p(t)$ is periodic with period $T$, i.e. $x_p(0) = x_p(T)$. By the mean value theorem, we know that there is a time in between where $\dot{x}_p$ vanishes, i.e. there exists $t_* \in (0, T)$ such that $\dot{x}_p(t_*) = 0$.
Since $x_p$ is a solution of $\dot{x} = f(x)$, we know that the $x_p$-value at $t = t_*$ is a fixed point of $f$, because $\dot{x}_p(t_*) = 0 = f(x_p(t_*))$. However, we also know that the constant solution $x_c = x_p(t_*)$ solves the ODE $\dot{x} = f(x)$, because $\dot{x}_c = 0 = f(x_c)$.
To summarize, we have two solutions of the ODE $\dot{x} = f(x)$ which pass through the point $(t_*,x_p(t_*))$: the periodic solution $x_p(t)$ and the constant solution $x_c$. That's a bit of a strange situation, because that means that if we take $(t_*,x_p(t_*))$ as the initial condition for our ODE $\dot{x} = f(x)$, that we can make a choice between two trajectories. In other words, this would imply non-uniqueness. If $f$ is locally Lipschitz, then by the Picard-Lindelöf theorem we know that this cannot happen, i.e. that $x_c$ and $x_p$ must be equal. So, if $f$ is locally Lipschitz, every periodic solution to $\dot{x} = f(x)$ is necessarily constant (and therefore not very exciting).
Interestingly enough, it turns out that you don't even need to assume that $f$ is locally Lipschitz to prove that nonconstant periodic solutions to $\dot{x} = f(x)$ do not exist. For more information, see @hardmath's answer to this question.