Solution of $\frac{dy}{dx}=\frac{1}{xy(x^2 \sin y^2+1)}$ Find the solution of following differential equation:

$$\frac{dy}{dx}=\frac{1}{xy(x^2 \sin (y^2)+1)}$$

Could someone hint me something to get through this problem?
 A: As @Nicholas Stull suggested, let $u=y^2$ so that
$$\frac{du}{dx}=\frac2{x\left(x^2\sin(y^2)+1\right)}$$
Then rearrange to Bernoulli's differential equation
$$\frac{dx}{du}-\frac12x=\frac12x^3\sin u$$
We may substitute $x=v^n$, and if $n-1=3n$, the dependence on the dependent variable on the right hand side will vanish. So $n=-\frac12$, $x=v^{-\frac12}$, and
$$\frac{dv}{du}+v=-\sin u$$
This is a first order linear differential equation with integrating factor
$$\mu=e^{\int1du}=e^u$$
So
$$\frac d{du}\left(e^uv\right)=e^u\frac{dv}{du}+e^uv=-e^u\sin u$$
Then
$$e^uv=-\frac{(e^u\sin u-e^u\cos u)}2+C$$
$$v=-\frac12\sin u+\frac12\cos u+Ce^{-u}$$
In terms of $x$ and $y$,
$$\frac1{x^2}=-\frac12\sin(y^2)+\frac12\cos(y^2)+Ce^{-y^2}$$
Which agrees nicely with the Wolfram|Alpha solution.
A: This is an exact differential equation. You can solve it using the method described at the Wiki page. Note that the obtain solution will be implicit, i.e. of the form $F(x,y) = c$. In this particular case, it is not possible to write this in the form $y = f(x)$. However, you can write $x$ in terms of $y$.
