= Solution
Here $M_y-N_x=(6xy+2)-(4xy+1)=2xy+1$, whereas $N=x(2xy+1)$. Thus the ratio in (b) is $1/x$, and one may choose the nonzero <integrating factor> $\boxed{\mu(x)=x}$ on either half-line. A negative constant multiple on a half-line is harmless.
The multiplied form is
$$
(3x^2y^2+2xy)\,dx+(2x^3y+x^2)\,dy
=d(x^3y^2+x^2y).
$$
Hence $x^3y^2+x^2y=C$, a quadratic equation for $y$. Solving it gives
$$
\boxed{y(x)=\frac{-1\pm\sqrt{1+4C/x}}{2x}},\qquad x\ne0.
$$
Choose a real branch with positive discriminant to obtain a differentiable graph; a zero discriminant produces a vertical tangent. Although the ratio derivation initially cancels $2xy+1$, the displayed exact differential verifies the multiplying factor directly wherever $x\ne0$. At $C=0$ the branches are $y=0$ and $y=-1/x$. Only $y=0$ has a finite smooth extension across $x=0$, where the original equation requires $y=0$.
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