The polynomial $P(X,Y)=XY-(X+Y)(X^2+Y^2)$ is irreducible in $K[X,Y]$, as a sum of two homogenous forms of degree 2 and 3 ($K$ is supposed to be algebraically closed). To look at the irreducibility in $K[[X,Y]]$, I first worked in $K[[X]][Z]$, where $Y=X^2Z$.
The polynomial becomes $Q(X,Z)=X^3(X^3Z^3+X^2Z^2+(X-1)Z+1)$. Using Hensel's Lemma in $K[[X]]$, I found that the polynomial $F(Z)=X^3Z^3+X^2Z^2+(X-1)Z+1$ has $1$ as a root modulo (X), and, with $F'(Z)=3X^3Z^2+2X^2Z+X-1$, $F'(1)=3X^3+2X^2+X-1$ is a unit in $K[[X]]$. Therefore, there is a root to $F$ looking like $Z=1+XH(X)$ where $H\in K[[X]]$. Then we can write $F(Z)=(Z-1-XH(X))(X-1+X^2(XZ^2+(1+XG(X))Z+T(X)))$ where $G, T\in K[[X]]$.
Going back to the initial polynomial $P$, we found that $P(X,Y)=(Y-X^2+X^2H(X))(X^2-X+Y^2+XY+X^2G(X)Y+X^3T(X))$ where $H,G,T \in K[[X]]$. Since the two factors of $P$ are not invertible, and $K[[X,Y]]$ is a UFD, we can deduce that $P$ is reducible.
After two unsucessful approaches, this is the easiest one I found (and brute force looks too difficult). Is there a less pedestrian/ad hoc method for this ?
