If $y=f\left(\frac{2 x-1}{x^2+1}\right)$ and $f'(x)=\sin x^2$ then $\frac{d y}{d x}$ is:

$\frac{2\left(1+x-x^2\right)}{\left(x^2+1\right)^2} \sin \left(\frac{2 x-1}{x^2+1}\right)^2$

Let $\frac{2 x-1}{x^2+1}=z \Rightarrow y=f(z)$

∴  $\frac{d y}{d x}=f'(z) . \frac{d z}{d x}$

$=\sin z^2 . \frac{d z}{d x}\left(∵ f'(z)=\sin z^2\right)$

$=\sin \left(\frac{2 x-1}{x^2+1}\right)^2 \frac{d}{d x}\left(\frac{2 x-1}{x^2+1}\right)$

$=\sin \left(\frac{2 x-1}{x^2+1}\right)^2 \frac{2\left(1+x-x^2\right)}{\left(x^2+1\right)^2}$

Hence (3) is correct answer.