 Solution

Writing the two coefficients as $M,N$, their mixed derivatives satisfy $M_y=2xy=N_x$, so the smooth form is exact on ${\mathbb{R}}^2$. Integrating the first coefficient with respect to $x$ gives $f=2x^2+x^2{y}^2/2+h(y)$. Matching $f_y=y+x^{2}y$ gives $h^{\prime }(y)=y$. One potential is thereforeThe integral curves of $df=0$ are its level curves. Absorbing a factor of two into the constant giveson real graph branches. The initial condition fixes $C=12$ and selects the positive branch:This is the maximal smooth graph interval containing $x=1$; at either endpoint the level curve has a vertical tangent, so $y^{\prime }$ cannot remain finite.