= Solution
Write the differential form as $M\,dx+N\,dy$, with the coefficients as printed. A <monomial integrating factor> can be found from $\mu=x^\alpha y^\beta$. The exactness condition $\partial_y(\mu M)=\partial_x(\mu N)$ gives
$$
2(\beta+2)=-(\alpha+2),\qquad -(\beta+1)=2(\alpha+1).
$$
Solving yields $\alpha=0,\beta=-3$. Thus, on $y\ne0$,
$$
\boxed{\mu(y)=y^{-3}}.
$$
After multiplication, the differential form is
$$
\left(\frac{2x}{y}-\frac1{y^2}\right)dx+
\left(\frac{2x}{y^3}-\frac{x^2}{y^2}\right)dy
=d\left(\frac{x^2}{y}-\frac{x}{y^2}\right).
$$
Hence the implicit solutions are
$$
\boxed{\frac{x^2}{y}-\frac{x}{y^2}=C\quad(y\ne0)}.
$$
The branch $\boxed{y=0}$ was excluded by the <integrating factor> but satisfies the original equation and must be restored. On any graph branch with $y\ne0$, differentiation of the boxed <first integral> recovers the original <differential equation>.
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