Given curve is
$y = (1 + x)^y + \sin^{-1} (\sin^2\, x)$
On differentiating w.r.t x, we get
$\frac{dy}{dx} = \left( 1+x\right)^{y} \left[ \frac{y}{1+x} + \log\left(1+x\right) \frac{dy}{dx}\right]$
$+ \frac{2\sin x \cos x}{\sqrt{1- \sin^{4} x }}$
$ \Rightarrow \left(\frac{dy}{dx}\right) _{at \left(0,1\right)} = 1$
$\left[\because \, at \, x = 0 , y = 1 \right] $
Slope of normal at $(x = 0) = -1$
$\therefore $ Equation of normal at $x = 0$ and $y = 1$ is
$y - 1 = -1 (x - 0) $
$\Rightarrow y - 1 = - x$
$\Rightarrow x + y = 1 $