Solution
= Solution
On the real domain $y>0$, $M=y(x^2+\ln y)$ and $N=x$ have $M_y=x^2+\ln y+1$ and $N_x=1$, so this <differential one-form> is not exact. An <integrating factor for a differential one-form> is $\mu=1/y$, which is nonzero on this domain. Multiplication gives
$$
(x^2+\ln y)\,dx+\frac{x}{y}\,dy=d\left(\frac{x^3}{3}+x\ln y\right).
$$
Hence the implicit solution is
$$
\boxed{\frac{x^3}{3}+x\ln y=C,\qquad y>0.}
$$