Dividing by $\cos^2y$ through out

$\dfrac 1{\cos^2y}\dfrac {dy}{dx}+\dfrac {x^2\sin^2y}{\cos^2y}=x^3\\ \sec^2y\dfrac {dy}{dx}+\dfrac {x^2\sin y\cos y}{\cos^2y}=x^3\\ \therefore \sec^2y\dfrac {dy}{dx}+2x\tan y=x^3......(1)\\ Let \space v=\tan y\\ \therefore \dfrac {dv}{dx}=\sec^2y\dfrac {dy}{dx}$

Substituting these values in eq.(1) we get,

$\dfrac {dv}{dx}+2xv=x^3$

Which is same as that of

$\dfrac {dv}{dx} + Pv=Q \\ where\space P=2x \space \& \space Q=x^3 \\ \therefore I.F=e^{\int Pdx}=e^{\int 2x\space dx}\\ =e^{x^2}\\ \therefore Solution \space is \\ v.I.F=\int Q.IF \space dx\\ v.e^{x^2}=\int x^3e^{x^2} dx...... (2)\\ Let \space x^2=U\\ \therefore 2x\space dx=du\\ \therefore \text{eq(2) becomes}\\ v.e^u=\int U\space e^U \space du\\ =\dfrac 12[Ue^U-e^U] + c\\ 2v.e^U=ue^u-e^u+2c\\ \therefore 2v=U-1+2ce^{-U}$

Resubstituting value of U and V

$\therefore 2\tan y=x^2-1+2ce^{-x^2}$