First, let's show the given DE $x^{2}y^{3} + x(1+y^{2})y' = 0$ is not exact.

Define $M(x,y)=x^{2}y^{3}$, $N(x,y)=x(1+y^{2})$

$$M_y = \frac{\partial}{\partial y}[x^{2}y^{3}] = 3x^{2}y^{2}$$ $$N_x = \frac{\partial}{\partial x}[x(1+y^{2})] = 1+y^{2} $$

Since $3x^{2}y^{2} ≠ 1+y^{2}$, this implies the given DE is not exact.

Now, let's show that the given DE multiplied by the integrating factor $\mu(x,y) = \frac{1}{xy^{3}}$ is exact.

That is to show $$\frac{1}{xy^{3}}x^{2}y^{3} + \frac{1}{xy^{3}}x(1+y^{2})y' = x + (y^{-3}+y^{-1})y'= 0$$ is exact.

Define $M'(x,y) = x$, $N'(x,y) = y^{-3}+y^{-1}$

Since

$$M'_y = \frac{\partial}{\partial y}(x) = 0 $$ $$N'_x = \frac{\partial}{\partial x}[y^{-3}+y^{-1}] = 0 $$

By theorem in the book, we can conclude that $x + (y^{-3}+y^{-1})y'= 0$ is exact.

Thus, we know there exists a function $\phi(x,y)=C$ which satisfies the given DE.

Also,

$$ \frac{\partial \phi}{\partial x} = x $$ $$ \frac{\partial\phi}{\partial y} = y^{-3}+y^{-1}$$

Integrate $\frac{\partial \phi}{\partial x} = x $ with respect to $x$ we have

$$\phi(x,y) = \frac{1}{2}x^{2} + g(y)$$

Take derivative on both sides with respect to $y$ we get

$$\frac{\partial\phi}{\partial y} = g'(y)$$

Since we know that $\frac{\partial\phi}{\partial y} = y^{-3}+y^{-1}$

Then $g'(y) = y^{-3}+y^{-1}$

Integrate with respect to $y$ we have

$g(y) = -\frac{1}{2}y^{-2} + ln|y| + C$

Altogether, we have $\phi(x,y) = \frac{1}{2}x^{2} -\frac{1}{2}y^{-2} + ln|y| = C $, which means

$$\frac{1}{2}x^{2} -\frac{1}{2}y^{-2} + ln|y| = C$$

is the general solution to the given DE.

Besides, notice that the constant function $y(x)=0$ $\forall x$ is also a solution to the given DE.