Solution
= Solution
The <chain rule> gives $dF(x,y(x))/dx=F_x+F_yy'$. Equality with $P+Qy'$ for every $y(x)$ therefore requires $F_x=P$ and $F_y=Q$, which implies $P_y=Q_x$. Conversely, on the plane this compatibility condition makes $P\,dx+Q\,dy$ an <exact differential>, so a potential $F$ exists.
Here $P=4x^3+3y$, $Q=2y+3x$, and $P_y=Q_x=3$. Integrating gives $F=x^4+3xy+y^2$, hence
$$
\boxed{x^4+3xy+y^2=C.}
$$