Let's prove it's neither Noetherian nor Artinian.
To each $A \subseteq \mathbb{R}$, we can assign a set $\varphi(A) \subseteq \mathbb{R}^\mathbb{R}$ as follows:
$$\varphi(A) = \{f \in \mathbb{R}^\mathbb{R} : \forall x \in A, f(x) = 0\}.$$
Exercise 0. Show that $\varphi(A)$ is an ideal for all sets $A \subseteq \mathbb{R}$.
Conclude that $$\varphi : \mathcal{P}(\mathbb{R}) \rightarrow \mathcal{P}(\mathbb{R}^\mathbb{R})$$ can be viewed as a function $$\mathcal{P}(\mathbb{R}) \rightarrow \mathrm{Ideal}(\mathbb{R}^\mathbb{R}).$$
Exercise 1. Show that $\varphi : \mathcal{P}(\mathbb{R}) \rightarrow \mathrm{Ideal}(\mathbb{R}^\mathbb{R})$ is injective and order-reversing.
Exercise 2. Find an order-reversing injection $\psi : \mathbb{Z} \rightarrow \mathcal{P}(\mathbb{R})$.
Conclude that $\varphi \circ \psi : \mathbb{Z} \rightarrow \mathrm{Ideal}(\mathbb{R}^\mathbb{R})$ is an order-preserving injection.
Conclude that $\mathbb{R}^\mathbb{R}$ is neither Artinian nor Noetherian.