Solution (a) Let $\ M=2x+3x^{2}y$ and $N=x^{3}+2y-3y^{2}.$ Then $$\frac{\partial M}{\partial y}=3x^{2}\qquad \hbox{and}\qquad \frac{\partial N }{\partial x}=3x^{2}$$

Since $\dfrac{\partial M}{\partial y}=\dfrac{\partial N}{\partial x}$ for all $x$ and $y$, the differential equation (1) is exact over the $xy$-plane.

(b) Since the differential equation $(2x+3x^{2}y)\,dx +(x^{3}+2y-3y^{2})\,dy=0$ is exact, there is a function $z=f( x,y)$ so that $$\frac{\partial f}{\partial x}=M=2x+3x^{2}y\qquad \hbox{and}\qquad \frac{\partial f}{\partial y}=N=x^{3}+2y-3y^{2}$$

We find $f$ by integrating $\dfrac{\partial f}{\partial x}=M=2x+3x^{2}y$ partially with respect to $x$ (holding $y$ constant). Then $$\begin{equation*} f( x,y) =\int ( 2x+3x^{2}y)\,dx=x^{2}+x^{3}y+B(y) \tag{2} \end{equation*}$$

where the constant of integration is a function $B$ of $y$ alone, which is as yet unknown. To determine $B(y)$, we use the fact that $f$ must also satisfy $\dfrac{\partial f}{\partial y}=N=x^{3}+2y-3y^{2}$. Then from (2), $$\begin{equation*} \frac{\partial f}{\partial y}=\dfrac{\partial }{\partial y}[ x^{2}+x^{3}y+B(y)] =x^{3}+\dfrac{\partial }{\partial y}B(y)\qquad \hbox{and}\qquad \frac{\partial f}{\partial y}=x^{3}+2y-3y^{2} \end{equation*}$$

Equating the two expressions, we have $$\begin{eqnarray*} x^{3}+\dfrac{\partial }{\partial y}B(y) &=&x^{3}+2y-3y^{2} \\[4pt] \dfrac{\it dB}{dy} &=&2y-3y^{2}\qquad \color{#0066A7}{{B\hbox{ is a function of }y\hbox{ alone; }\quad \dfrac{\partial B}{\partial y}=\dfrac{dB}{dy}.}} \end{eqnarray*}$$

Now we integrate $\dfrac{\it dB}{dy}=2y-3y^{2}$ with respect to $y.$ $$\begin{equation*} B(y) =\int ( 2y-3y^{2})\,dy=y^{2}-y^{3}+C \end{equation*}$$

Substituting for $B(y)$ in (2), we obtain $$\begin{equation*} f(x,y)=x^{2}+x^{3}y+y^{2}-y^{3}+C \end{equation*}$$

The general solution of the differential equation is $$\begin{equation*} x^{2}+x^{3}y+y^{2}-y^{3}+C=0 \end{equation*}$$

where $C$ is a constant.