Step 1: Confirm exactness: with $M=2x+y+1$ and $N=x+2y+1$, $M_y=N_x=1$, so a potential function $F$ exists.
Step 2: Recover $F$ using the line-integral formula \[F(x,y)=\int_0^x M(t,0)\,dt+\int_0^y N(x,t)\,dt.\]
Step 3: Compute the first integral: $M(t,0)=2t+1$, so \[\int_0^x (2t+1)\,dt=x^2+x.\]
Step 4: Compute the second integral: $N(x,t)=x+2t+1$, so \[\int_0^y (x+2t+1)\,dt=xy+y^2+y.\]
Step 5: Adding gives the same implicit solution $x^2+xy+y^2+x+y=C$. Write the quadratic part in matrix form $\begin{pmatrix}x & y\end{pmatrix}\begin{pmatrix}1 & 0.5\\ 0.5 & 1\end{pmatrix}\begin{pmatrix}x\\ y\end{pmatrix}$.
Step 6: The eigenvalues of $\begin{pmatrix}1 & 0.5\\ 0.5 & 1\end{pmatrix}$ are $1.5$ and $0.5$, both positive, so the quadratic form is positive definite. Every level curve $x^2+xy+y^2+x+y=C$ is therefore a bounded, real ellipse.
\[\boxed{\text{Family of ellipses}}\]