Your question has already been answered by monkeymaths, but here is some more information.

The proof that every planar graph can be partitioned into *three* induced forests is actually quite easy. Let $G$ be a planar graph and $v$ be a vertex of degree at most $5$. By induction, $G-v$ can be partitioned into three induced forests. Since $v$ has degree at most $5$, one of these forests $F$ contains at most one neighbour of $v$. Thus, we can simply add $v$ to $F$.

Regarding the question of the largest induced forest of a planar graph (raised by monkeymaths), we can use the following theorem of Borodin.

**Theorem** (Borodin). Every planar graph has an acyclic $5$-colouring.

Here, an *acyclic colouring* is a proper colouring such that every $2$-coloured subgraph is acyclic (note that this is a different definition than in the original question, but is quite standard) . By taking the two largest colour classes of an acyclic $5$-colouring, we get that every $n$-vertex planar graph has an induced forest with at least $\frac{2n}{5}$ vertices. As far as I know, this is still the best lower bound. There are however improved bounds for triangle-free planar graphs. For example, Dross, Montassier, and Pinlou recently proved that every $n$-vertex triangle-free planar graph contains an induced forest with at least $\frac{6n+7}{11}$ vertices.