Answer :
[tex]\dfrac{\mathrm dy}{\mathrm dx}=y^2\sin(x^2)[/tex]
is separable as
[tex]\dfrac{\mathrm dy}{y^2}=\sin(x^2)\,\mathrm dx[/tex]
Integrating both sides gives
[tex]\displaystyle\int\frac{\mathrm dy}{y^2}=\int\sin(x^2)\,\mathrm dx[/tex]
The integral on the right unfortunately has no elementary result (i.e. not in terms of Fresnel integrals). But we are given an initial value, so we can use the fundamental theorem of calculus to find a solution in terms of an integral.
Recall that
[tex]\displaystyle\int_a^bf(x)\,\mathrm dx=F(b)-F(a)[/tex]
where [tex]f(x)=\frac{\mathrm dF}{\mathrm dx}[/tex].
Let [tex]F[/tex] be the antiderivative of [tex]\sin(x^2)[/tex]. The theorem allows us to write
[tex]\displaystyle F(x)=F(-4)+\int_{-4}^x\sin(u^2)\,\mathrm du[/tex]
So the solution to the ODE is (upon integrating both sides)
[tex]\displaystyle-\frac1y=\frac16+\int_{-4}^x\sin(u^2)\,\mathrm du[/tex]
Just to confirm this is valid, differentiating both sides gives
[tex]\dfrac1{y^2}\dfrac{\mathrm dy}{\mathrm dx}=\sin(x^2)\implies\dfrac{\mathrm dy}{\mathrm dx}=y^2\sin(x^2)[/tex]
so all is well.
We can go on to solve for [tex]y[/tex] directly by taking the reciprocal of both sides, ending up with
[tex]\boxed{y(x)=-\dfrac1{\frac16+\int\limits_{-4}^x\sin(u^2)\,\mathrm du}}[/tex]