= Solution
For continuously differentiable coefficients on a simply connected domain, the exactness condition is \b[$P_y=Q_x$]. Locally the same condition suffices without a global topological assumption. An <exact first-order ordinary differential equation> has a potential $F$ with $F_x=P$ and $F_y=Q$, so $dF/dx=P+Qy'=0$ along a solution. Thus its <implicit solution> is $F(x,y)=C$. One local construction, on a rectangle about $(x_0,y_0)$, is
$$
F(x,y)=\int_{x_0}^xP(s,y_0)\,ds+\int_{y_0}^yQ(x,t)\,dt.
$$
The equality of cross derivatives verifies both required derivatives; on a simply connected domain one may use the corresponding path-independent integral of the <exact differential form>.
Here $P=4x+3y$, $Q=3x+3y^2$ and $P_y=Q_x=3$. Integrating $P$ with respect to $x$ gives $F=2x^2+3xy+K(y)$. Matching $F_y=Q$ gives $K'=3y^2$, so take $K=y^3$. The initial value sets $C=F(1,2)=16$. Therefore
$$
\boxed{2x^2+3xy+y^3=16.}
$$
This is the requested explicit relation between the coordinates. Since $F_y(1,2)=15\ne0$, the <implicit function theorem> gives a unique local graph through the initial point, with slope $-10/15=-2/3$.
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