A solution of the differential equation 

$\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{xy\left[x^2\sin y^2+1\right]}$ is

                         ( C is an arbitrary constant)

The given differential equation can be written 

as $\frac{\mathrm{d}x}{\mathrm{d}y}=xy\left[x^2\sin y^2+1\right]$

$\Rightarrow \frac{1}{x^3}\frac{\mathrm{d}x}{\mathrm{d}y}-\frac{1}{x^2}y=y\sin y^2$ . This equation is reducible to

 linear equation, so putting $-1/x^2=u$ the last equation can

 be written as $\frac{\mathrm{d}u}{\mathrm{d}y}+2uy=2y \sin y^2$

The integrating factor of this equation is $e^{y^2}$. So required solution is

$\begin{array}{l}u{e}^{y^2}=\int 2y\sin y^2\cdot e^{y^2}\mathrm{d}y+C\\ =\int (\sin t)\cdot e^t\mathrm{d}t+C\left(t=y^2\right)\\ =(1/2)e^{y^2}\left(\sin y^2-\cos y^2\right)+C\\ \Rightarrow 2u=\left(\sin y^2-\cos y^2\right)+C{e}^{-y^2}\\ \Rightarrow 2=x^2\left[\cos y^2-\sin y^2-2C{e}^{-y^2}\right].\end{array}$