The equation of the normal to the curve $y=(1+x)^y+\sin ^{-1}\left(\sin ^2 x\right)$ at x = 0, is

$x+y=1$

We have,

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

When x = 0, we have y = 1

Differentiating (i) w.r.t. x, we get

$\frac{d y}{d x}=(1+x)^y\left\{\frac{d y}{d x} \log (1+x)+\frac{y}{1+x}\right\}+\frac{\sin 2 x}{\sqrt{1-\sin ^4 x}}$

$\Rightarrow \left(\frac{d y}{d x}\right)_{(0,1)}=1$

So, the equation of the normal at (0, 1) is

$y-1=-1(x-0) \Rightarrow x+y=1$