We have,
$f(x)=\tan ^{-1}\left(\frac{1}{\sin ^2 x+\sin x+1}\right)$ $+\tan ^{-1}\left(\frac{1}{\sin ^2 x+3 \sin x+3}\right)$ $+\tan ^{-1}\left(\frac{1}{\sin ^2 x+5 \sin x+7}\right)+\ldots+$ upto 10 terms
$=\tan ^{-1}\left(\frac{(\sin x+1)-(\sin x+0)}{1+(\sin x+0)(\sin x+1)}\right)$ $+\tan ^{-1}\left(\frac{(\sin x+2)-(\sin x+1)}{1+(\sin x+1)(\sin x+2)}\right)$ $+\tan ^{-1}\left(\frac{(\sin x+3)-(\sin x+2)}{1+(\sin x+2)(\sin x+3)}\right)+\ldots +$ upto 10 terms
$=\tan ^{-1}(\sin x+1)-\tan ^{-1}(\sin x)$ $+\tan ^{-1}(\sin x+2)-\tan ^{-1}(\sin x+1)$ $+\tan ^{-1}(\sin x+3)-\tan ^{-1}(\sin x+2)+\ldots$ + upto 10 terms
$=\tan ^{-1}(\sin x+10)-\tan ^{-1}(\sin x)$
$\therefore f^{\prime}(x)=\frac{\cos x}{1+(\sin x+10)^2}-\frac{\cos x}{1+(\sin x)^2}$
$\therefore f^{\prime}(0)=\frac{1}{1+(0+10)^2}-\frac{1}{1+0}$
$=\frac{1}{101}-1=\frac{1-101}{101}=\frac{-100}{101}$