A function y = f(x)  satisfies  the differential equation  $f\left(x\right).\left(\sin 2x\right)-\cos x+(1+{\sin }^{2}x)f\text{'}\left(x\right)=0$ with initial condition y (0) =0. the value of  $f(\pi /6)$ is equal to

$$\begin{gathered} y\sin 2x-\cos x+(1+{\sin }^{2}x)\frac{dy}{dx}=0 wherey=f\left(x\right) \\ \frac{dy}{dx}+\left(\frac{\sin 2x}{1+{\sin }^{2}x}\right)y=\frac{\cos x}{1+{\sin }^{2}x} \\ IF=e^{\int \frac{\sin 2x}{1+{\sin }^{2}x}dx}=e^{\int \frac{dt}{t}}=e^{In(1+{\sin }^{2}x)} \\ =1+{\sin }^{2}x(byputting1+{\sin }^{2}x=t) \\ y(1+{\sin }^{2}x)=\int \cos xdx \\ y(1+{\sin }^{2}x)=\sin x+C;y\left(0\right)=0\Rightarrow C=0 \end{gathered}$$

Therefore,     $y=\frac{\sin x}{1+{\sin }^{2}x};y\left(\frac{\pi }{6}\right)=\frac{2}{5}$
Hence, (d) is the correct answer.